#	User	Rating
1	tourist	3985
2	jiangly	3814
3	jqdai0815	3682
4	Benq	3529
5	orzdevinwang	3526
6	ksun48	3517
7	Radewoosh	3410
8	hos.lyric	3399
9	ecnerwala	3392
9	Um_nik	3392

#	User	Contrib.
1	cry	169
2	maomao90	162
2	Um_nik	162
4	atcoder_official	161
5	djm03178	158
6	-is-this-fft-	157
7	adamant	155
8	awoo	154
8	Dominater069	154
10	luogu_official	150

Author: Anadi

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

LL getR(LL a){
	while(a % 2 == 0)
		a /= 2;
	return a;
}

void solve(){
	LL a, b;
	scanf("%lld %lld", &a, &b);
	if(a > b)	swap(a, b);
	
	LL r = getR(a);
	if(getR(b) != r){
		puts("-1");
		return;
	}
	
	int ans = 0;
	b /= a;
	
	while(b >= 8)
		b /= 8, ++ans;
	if(b > 1)	++ans;
	printf("%d\n", ans);
}

int main(){
	int quest;
	scanf("%d", &quest);
	
	while(quest--)
		solve();
	return 0;
}

1362B - Johnny and His Hobbies

Authors: Anadi and Okrut

Tutorial

Solution

//O(n * maxA) solution

#include <bits/stdc++.h>

using namespace std;

const int N = 1025;

int n;
int in[N];
bool is[N];

bool check(int k){
	for(int i = 1; i <= n; ++i)
		if(!is[in[i] ^ k])
			return false;
	return true;
}

void solve(){
	for(int i = 0; i < N; ++i)
		is[i] = false;
	
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i){
		scanf("%d", &in[i]);
		is[in[i]] = true;
	}
	
	for(int k = 1; k < 1024; ++k)
		if(check(k)){
			printf("%d\n", k);
			return;
		}
	
	puts("-1");
}

int main(){
	int cases;
	scanf("%d", &cases);
	
	while(cases--)
		solve();
	return 0;
}

1362C - Johnny and Another Rating Drop

Author: MicGor

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

void solve(){
	LL a;
	scanf("%lld", &a);
	
	LL ans = 0;
	for(int i = 0; i < 60; ++i)
		if(a & (1LL << i))
			ans += (1LL << (i + 1)) - 1;
	printf("%lld\n", ans);
}

int main(){
	int quest;
	scanf("%d", &quest);
	
	while(quest--)
		solve();
	return 0;
}

1361A - Johnny and Contribution

Author: Anadi

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 7;

int n, m;
int ans[N];

int last[N];
vector <int> G[N];
vector <int> in[N];

int main(){
	scanf("%d %d", &n, &m);
	for(int i = 1; i <= m; ++i){
		int u, v;
		scanf("%d %d", &u, &v);
		
		G[u].push_back(v);
		G[v].push_back(u);
	}
	
	for(int i = 1; i <= n; ++i){
		int color;
		scanf("%d", &color);

		in[color].push_back(i);
		if(color > n){
			puts("-1");
			exit(0);
		}
	}
	
	vector <int> result;
	for(int i = 1; i <= n; ++i){
		for(auto u: in[i]){
			for(auto v: G[u])
				last[ans[v]] = u;
			
			ans[u] = 1;
			while(last[ans[u]] == u)
				++ans[u];
			
			if(ans[u] != i){
				puts("-1");
				exit(0);
			}
			
			result.push_back(u);
		}
	}
	
	for(int i = 0; i < n; ++i)
		printf("%d%c", result[i], i == n - 1 ? '\n' : ' ');
	return 0;
}

1361B - Johnny and Grandmaster

Author: Okrut

Tutorial

Solution

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const ll INF = 1e6+100;
const ll mod = ll(1e9)+7;

ll fexp (ll a, int e)
{
	ll ret = 1LL;
	while (e>0)
	{
		if (e%2==1) ret = ret * a % mod;
		a = a*a % mod;
		e/=2;
	}
	return ret;
}

int main ()
{
	int t;
	scanf ("%d", &t);
	for (int test = 0; test < t; test++)
	{
		int n;
		ll p;
		scanf ("%d %lld", &n, &p);
		vector <int> K(n);
		for (int &x: K) scanf ("%d", &x);
		sort(K.begin(), K.end());
		
		vector <int> Exp = K;
		
		ll currSum = 0;
		ll result = 0;
		bool infty = false;
		
		int prevExp = 1e6+10;
		while (!Exp.empty())
		{
			/* Update the currSum and result (multiply by a power of k) */
			int k = Exp.back();
			Exp.pop_back();
			int delta = prevExp - k;
			prevExp = k;
			
			result = result * fexp(p, delta) % mod;
			
			/* Multiplying by p 20 times will either 
			not change the currSum at all or make it bigger than n */
			for (int i=0; i<min(delta, 20); i++)
			{
				currSum *= p;
				if (currSum > INF) infty = true;
			}
			
			/* Process all the elements equal p^k */
			while (!K.empty() && K.back()==k)
			{
				K.pop_back();
				if (infty || currSum > 0)
				{
					currSum--;
					result+=mod - 1;
				}
				else
				{
					currSum++;
					result++;
				}
				result%=mod;
			}
		}
		
		result = result * fexp(p, prevExp) % mod;
		
		printf ("%lld\n", result % mod);
	}
}

1361C - Johnny and Megan's Necklace

Author: Okrut

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

#define st first
#define nd second
#define PII pair <int, int>

const int N = 1 << 20;

int n;
int part[N][2];

bool vis[N];
vector <int> ans;
vector <PII> G[N];

//mark whole component as visited
void dfs(int u){
	vis[u] = true;
	for(auto v: G[u])
		if(!vis[v.st])
			dfs(v.st);
}

//check if the graph for a given Mask is correct
bool check(int Mask){
	for(int i = 0; i <= Mask; ++i)
		G[i].clear(), vis[i] = false;

	for(int i = 1; i <= n; ++i){
		int u = part[i][0] & Mask;
		int v = part[i][1] & Mask;
		
		G[u].push_back({v, 2 * i - 1});
		G[v].push_back({u, 2 * i - 2});
	}
	
	//calculate number of components and check if all degrees are even
	int comps = 0;
	for(int i = 0; i <= Mask; ++i){
		if(G[i].size() & 1)
			return false;
		
		if(!vis[i] && G[i].size() > 0){
			++comps;
			dfs(i);
		}
	}
	
	return comps == 1;
}

//find euler cycle
void go(int u, int prv = -1){
	while(G[u].size()){
		auto e = G[u].back();
		G[u].pop_back();

		if(vis[e.nd / 2])
			continue;
		
		vis[e.nd / 2] = true;
		go(e.st, e.nd);
	}
	
	if(prv != -1){
		ans.push_back(prv);
		ans.push_back(prv ^ 1);
	}
}

//restore the sequence corresponding to the result for given mask
//the result is already checked so the graph is built
void restore(int Mask){
	for(int i = 0; i <= n; ++i)
		vis[i] = false;
	
	for(int i = 0; i <= Mask; ++i)
		if(G[i].size()){
			go(i);
			break;
		}
	
	//cycle is reversed but thats fine
	for(int i = 0; i < n + n; ++i)
		printf("%d%c", ans[i] + 1, " \n"[i + 1 == n + n]);
}

int main(){
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i)
		scanf("%d %d", &part[i][0], &part[i][1]);

	for(int i = 20; i >= 0; --i)
		if(check((1 << i) - 1)){
			printf("%d\n", i);
			restore((1 << i) - 1);
			exit(0);
		}
	
	assert(false);
	return 0;
}

1361D - Johnny and James

Author: Okrut

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

typedef long double T;

#define st first
#define nd second
#define PII pair <int, int>

const int N = 1e6 + 7;

int n, m, k;
vector <T> arms[N];

void read(){
	map <PII, int> M;

	scanf("%d %d", &n, &k);	
	for(int i = 1; i <= n; ++i){
		int x, y;
		scanf("%d %d", &x, &y);
		
		if(x == 0 && y == 0)
			continue;
		
		int d = __gcd(abs(x), abs(y));
		PII my_id = {x / d, y / d};
		
		if(!M.count(my_id))
			M[my_id] = ++m;
		
		auto id = M[my_id];
		arms[id].push_back(sqrtl((T)x * x + (T)y * y));
	}
	
	for(int i = 1; i <= m; ++i){
		sort(arms[i].begin(), arms[i].end());
		reverse(arms[i].begin(), arms[i].end());
	}
}

T solve_center(){
	T ans = 0;
	vector <T> vals = {0};

	for(int i = 1; i <= m; ++i){
		int it = 0;
		for(auto v: arms[i]){
			++it;
			vals.push_back(v * (k - it - it + 1));
		}
	}
	
	sort(vals.begin(), vals.end());
	reverse(vals.begin(), vals.end());

	for(int i = 0; i < k; ++i)
		ans += vals[i];
	return ans;
}

T solve_arm(int id){
	T ans = 0;
	for(int i = 1; i <= m; ++i){
		if(i == id)
			continue;
		
		int it = 0;
		for(auto v: arms[i]){
			++it;
			ans += v * (k - it - it + 1);
		}
	}
	
	int half = k / 2;
	int first_half = k - half - (n - arms[id].size());
		
	for(int i = 0; i < half; ++i)
		ans += arms[id][i] * (k - i - i - 1);
	
	for(int i = 0; i < first_half; ++i){
		T val = arms[id][arms[id].size() - i - 1];
		ans += val * (k - half - half - 2 * first_half + 2 * i + 1);
	}
	
	return ans;
}

int main(){
	read();
	T ans = solve_center();
	if(k < n){
		for(int i = 1; i <= m; ++i)
			if(2 * (n - (int)arms[i].size()) <= k)
				ans = max(ans, solve_arm(i));
	}

	printf("%.10Lf\n", ans);
	return 0;
}

1361E - James and the Chase

Author: Anadi

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

const int N = 1e5 + 7;

int n, m;
bool bad[N];
vector <int> G[N];

int vis[N];
bool interesting;

int lvl[N];
int best[N];
int balance[N];

void clear(){
	for(int i = 1; i <= n; ++i){
		lvl[i] = 0;
		best[i] = 0;
		balance[i] = 0;

		G[i].clear();		
		bad[i] = false;
	}
}

void no_answer(){
	puts("-1");
	clear();
}

int getInt(int a = INT_MIN, int b = INT_MAX){
	static mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	return uniform_int_distribution <int> (a, b)(rng);
}

void dfs(int u){
	vis[u] = 1;
	for(auto v: G[u])
		if(vis[v] == 0)
			dfs(v);
		else if(vis[v] == 2)
			interesting = false;
	vis[u] = 2;
}

bool check(int r){
	for(int i = 1; i <= n; ++i)
		vis[i] = 0;
	interesting = true;

	dfs(r);
	return interesting;
}

int find_any(){
	int tests = 100;
	while(tests--){
		int r = getInt(1, n);
		if(check(r))
			return r;
	}
	
	return -1;
}

int find_bad(int u){
	vis[u] = 1;
	best[u] = u;
	for(auto v: G[u])
		if(vis[v] == 0){
			lvl[v] = lvl[u] + 1;
			balance[u] += find_bad(v);
			
			if(lvl[best[v]] < lvl[best[u]])
				best[u] = best[v];
		}
		else{
			balance[u]++;
			balance[v]--;
			
			if(lvl[v] < lvl[best[u]])
				best[u] = v;
		}
	
	if(balance[u] > 1)
		bad[u] = true;
	return balance[u];
}

void propagate_bad(int u){
	vis[u] = 1;
	if(!bad[u] && bad[best[u]])
		bad[u] = true;

	for(auto v: G[u])
		if(vis[v] == 0)
			propagate_bad(v);
}

vector <int> find_all(int r){
	for(int i = 1; i <= n; ++i)
		vis[i] = 0;
	vector <int> ans;

	find_bad(r);
	for(int i = 1; i <= n; ++i)
		vis[i] = 0;

	propagate_bad(r);
	for(int i = 1; i <= n; ++i)
		if(!bad[i])
			ans.push_back(i);
	return ans;
}

void solve(){
	scanf("%d %d", &n, &m);
	for(int i = 1; i <= m; ++i){
		int u, v;
		scanf("%d %d", &u, &v);		
		G[u].push_back(v);
	}
	
	int id = find_any();
	if(id == -1){
		no_answer();
		return;
	}
	
	auto ans = find_all(id);
	if(5 * ans.size() >= n){
		for(auto v: ans)
			printf("%d ", v);
		puts("");
		clear();
	}
	else
		no_answer();
}

int main(){
	int cases;
	scanf("%d", &cases);
	
	while(cases--)
		solve();
	return 0;
}

1361F - Johnny and New Toy

Author: Anadi

Tutorial

Solution

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

typedef long long ll;
typedef tree < int, null_type, less <int>, rb_tree_tag, tree_order_statistics_node_update > ordered_set;

const int N = 2e5 + 7;

//cartesian tree node
//while creating we calculate set of elements and number of inversions
struct node{
	int splitter = -1;
	long long invs = 0;	
	
	ordered_set S;
	node *left = nullptr, *right = nullptr;
	node(){}
};

int n;
int in[N];
int weight[N];

long long ans = 0;
node *root = nullptr;

void read(){
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i)
		scanf("%d", &in[i]);
	
	for(int i = 1; i < n; ++i)
		scanf("%d", &weight[i]);
}

//add/delete contribution from fiven node to result
void add_result(node *cur, int mt = 1){
	if(cur -> splitter == 0)
		return;
	ans += mt * min(cur -> invs, (ll)cur -> left -> S.size() * (ll)cur -> right -> S.size() - cur -> invs);
}

//Create cartesian tree
//Because expected depth is O(logn) we can use bruteforce to find the smallest weight
node* get_cartesian(int from, int to){
	node *ret = new node();
	if(from == to){
		ret -> S.insert(in[from]);
		return ret;
	}
	
	int id = from;
	for(int i = from + 1; i < to; ++i)
		if(weight[i] < weight[id])
			id = i;
	
	ret -> splitter = id;
	ret -> left = get_cartesian(from, id);
	ret -> right = get_cartesian(id + 1, to);
	
	//get elements from ordered set
	for(auto v: ret -> right -> S)
		ret -> S.insert(v);
	
	//calculate number of inversions
	for(auto v: ret -> left -> S){
		ret -> S.insert(v);
		ret -> invs += ret -> right -> S.order_of_key(v);
	}
	
	//add contribution to the result
	add_result(ret);
	return ret;
}

//depending on mt, removes/add element val on position p
void update_cart(int p, int val, int mt){
	node *cur = root;
	while(cur -> splitter != -1){
		add_result(cur, -1);
		if(mt == -1)
			cur -> S.erase(val);
		else
			cur -> S.insert(val);
		
		if(p <= cur -> splitter){
			cur -> invs += mt * (int)(cur -> right -> S.order_of_key(val));
			cur = cur -> left;
		}
		else{
			cur -> invs += mt * (int)(cur -> left -> S.size() - cur -> left -> S.order_of_key(val));
			cur = cur -> right;
		}
	}
	
	if(mt == -1)
		cur -> S.erase(val);
	else
		cur -> S.insert(val);	

	cur = root;
	while(cur -> splitter != -1){
		add_result(cur, 1);
		if(p <= cur -> splitter)
			cur = cur -> left;
		else
			cur = cur -> right;
	}
}

int main(){
	read();
	root = get_cartesian(1, n);
	
	int q;
	scanf("%d", &q);
	while(q--){
		int x, y;
		scanf("%d %d", &x, &y);
		
		if(x == y){
			printf("%lld\n", ans);
			continue;
		}
		
		//remove both numbers
		update_cart(x, in[x], -1);
		update_cart(y, in[y], -1);
		swap(in[x], in[y]);
		
		//add both numbers
		update_cart(x, in[x], 1);
		update_cart(y, in[y], 1);
		printf("%lld\n", ans);
	}
	
	return 0;
}

#include <iostream> #include<cmath> using namespace std; typedef long long ll; int main() { int t; cin>>t; while(t--){ ll n; ll res = 0; cin>>n; int highestBit = floor(log2(n)) + 1; for(int k = 0; k < highestBit; k++){ res += (n/(pow(2, k))); } cout<<res<<endl; } }

#include <cmath> #include <iostream> using namespace std; typedef long long ll; int main() { int t; cin >> t; while (t--) { ll n; ll res = 0; cin >> n; int highestBit = floor(log2(n)) + 1; for (int k = 0; k < highestBit; k++) { res += (n / (ll)pow(2, k)); } cout << res << endl; } }

import java.util.*; public class Rate_Drop { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int t = sc.nextInt(); while(t-- != 0) { long n = sc.nextLong(); long final_val = 0; while(n >= 2) { long log_val = (long)((Math.log(n) / Math.log(2))); final_val += func(log_val); n = n - (long)Math.pow(2, log_val); } if(n == 1) { final_val += 1; } System.out.println(final_val); } sc.close(); } public static long func(long n) { long prev = 0; for(long i = 0; i <= n; i++) { prev = prev * 2 + 1; } return prev; } }

#include<bits/stdc++.h> #define M 1000000007 #define pb push_back #define ll long long int using namespace std; ll binexp(ll x, ll y) { ll res = 1; while(y!=0) { if(y & 1) { res = (res*x)%M; } y=(y>>1); x = (x*x)%M; } return res; } int main() { int t; cin>>t; while(t--) { int n,p; cin>>n>>p; int k[n]; int i; for(i=0;i<n;i++) cin>>k[i]; sort(k,k+n); ll ans = binexp(p,k[n-1]); for(i=n-2;i>=0;i--) { ll curval = binexp(p,k[i]); if(ans==0) ans = curval; else ans=(ans+(M-curval))%M; } cout<<ans<<"\n"; } return 0; }

include<bits/stdc++.h>

typedef long long int ll; using namespace std;

int main()
{ ll t1; cin>>t1; while(t1--) { ll n; cin>>n; ll a[n]; set s1; for(ll i=0;i<n;i++) { cin>>a[i]; s1.insert(a[i]); } if(s1.size()==1) cout<<"0"<<endl; else {

ll f=1; for(ll i=1;i<=1023;i++) { f=0; for(ll j=0;j<n;j++) { if(s1.find((a[j]^(i)))==s1.end()) { f=1; break; } } if(f==0) { cout<<i<<endl; break; } } if(f==1) cout<<"-1"<<endl; } }

}

#include <iostream> #include <cmath> #include <cstring> using namespace std; int main(int argc, char** argv) { #ifdef LOCAL freopen(strcat(argv[0], ".in"), "ab+", stdin); #endif ios_base::sync_with_stdio(false); cin.tie(nullptr); int t; cin >> t; while (t--) { uint64_t a, b; cin >> a >> b; if (max(a, b) % min(a, b) == 0) { uint64_t x = max(a, b) / min(a, b); if (a == b) { cout << 0 << endl; } else if (log2(x) == ceil(log2(x))) { double p = log2(x); cout << ceil(p / 3) << endl; } else { cout << -1 << endl; } } else { cout << -1 << endl; } } return 0; }

- From Condition 2 : HERE -vertex 4 have immediate neighbors [2(color 2) and 3(color 2)] but there is no neighbor of color 1 -If we want to assign vertex 4 color 3 then there should be other vertex available of color 1 to 2

Comments (313)

Show archived | Write comment?

Ebiarat

5 years ago, # |

+11

Greate contest,thanks!

→ Reply

Pallove

5 years ago, # ^ |

+17

worst explanation of div2 D.

epsilon_573

← Rev. 4 →

Can't there be cycles in problem D? ( EDIT: Ignore I accidentally looked at D1 D and it said 'tree' so got confused. )

bobbilyking

i assume u mean div2D right?

i'm pretty sure it doesn't matter u greedy fill solutions starting from the lowest one after checking all the ways in which it could be invalid. ur not traversing the graph ur only checking the connections for each one

← Rev. 2 →

-7

I thought greedy won't work, so didn't implement it. Damn, feeling so dumb right now. I spent all my time looking for graph coloring problems as it was quite similar.

uchiha_asha

Problem D video Editorial: Solution

AQZZ

+15

Brute force O(1023 * n * log(n)) ~= O(n^2 * log(n)) solution passed in part B. I simply tested all values of k from 1 to 1023 until one worked.

82505984

vipinkumar17

sum of n over all test cases does not exceed 1024.

RomeoFantastik

+70

Video Editorials to A + B(Div.1) / D + E(Div. 2)

aadish

Actually they are D and E of Div 2

Oh that's right thanks!

kati-patang

In prob E: For converting an odd coefficient to even, why are we taking subarray starting from that point. We can take subset instead of subarray.

Mcayitel

Yes, you are correct but you will notice that we only need to consider subarray. Consider that you are at index i where you found odd occurrence.Now to make it even you need P occurrence from (i+1). Assume we have only have X occurrence at (i+1), so we need P*(P-x) occurrence from (i+2) and so on. Now,if I get my extra occurrence from a subset then whatever occurrence we are using from index j it must be first converted to in the form of (j+1) and so on. (indexs are consider after sorting array descendingly)

ishank162

-18

I edited my comment. Don't press <-- Rev.

RitikBaid

← Rev. 3 →

ur soln for D fails for the following tc:-

4 3

1 2

2 3

3 4

1 3 3 1

Looks like I really got lucky this round :D

xD..great editorial man!way to go!

_pastor_

great tutorial this time!!

what is that do_calc function is doing in your code for div2E/div1B
upd:- finally got that amazing solution thank you

AyanoGod

+18

seemingly simpler solution for problem C

ll res=0;
while(n){
    res+=n;
    n/=2;
}
cout<<res<<endl;

FireFeathers

+23

This was my C 2*n-__builtin_popcountll(2*n)

vickyvanshaj

Can you tell me why this works?

stronto9

We first note that the sum of the first 2^k values for any k is 2^(k+1) — 1. One can easily prove this via induction (or basically, noting that the sequence from 2^k + 1 to 2^(k+1) is identical except for the last number, which is increased by 1; then solving the recurrence).

Next, we use this identical property to our advantage. For a number n, we can write this number in binary as a sum of decreasing powers of 2. The key observation is, using the property, we can see that the first 2^m numbers after a number divisible by 2^n is identical to the first 2^m numbers, for n > m. This means that the sum of the first n numbers is the sum of the sum of the first 2^k numbers for each power of 2 in the decomposition.

Then, we group the powers of 2 in our sum together and the -1s together. Note that the powers of two are each 2* the original powers in the number; so they sum to 2^n. Next, the -1s appear for each power of 2 in the summation, aka the number of digits in binary. This is the __builtin_popcountll(n).

hosman

can someone explain this solution please, I'm having trouble understanding how this works.

Pranshu54

+27

If you make a sequence of bits from 1st place to 64th place:

1st-bit place sequence:0 1 0 1 0 1 0 1 0 1 0 1 0 1 0.....(n+1 terms)

2nd-bit place sequence:0 0 1 1 0 0 1 1 0 0 1 1 0 0 1.....(n+1 terms)

3rd-bit place sequence:0 0 0 0 1 1 1 1 0 0 0 0 1 1 1.....(n+1 terms) and so on... ..till 64th bit. For each bit sequence, the value added to the distance will be equal to the number of terms in sequence for which s[i]!=s[i-1](i.e. the number of terms which differ from their previous term).

1st-bit place sequence:Value added = n

2nd-bit place sequence:Value added = n/2

3rd-bit place sequence:Value added = n/4 and so on... ..till 64th bit.

Final answer = n + n/2 + n/4 + n/8 + ......

Hope this helps you to understand better.

namit2saxena

Hello! Thank you for the explanation. I have a doubt in mind regarding choosing n, n/2...

If our given number is 8 then, for 2nd LSB, our values change after every 2 numbers. The numbers are total (8-0+1)/2 = 4 from 0 to 8. So, there is a bit difference of 4. Whereas, your solution directly gives 8/2 = 4.

If our n = 13 then, for 2nd LSB we have (13-0+1)/2 = 7 which is odd. That means, that our n, and n-1 had 0 in their 2nd LSB. So, we should subtract one from 7 which should give 6. Your solution gives directly 13/2 = 6.

Did you choose n/2, n/4,.. instead of the other way that I have mentioned above because it works in all the cases and requires less computation or am I missing something?

Thank you!

I am not able to understand on what logic are you deciding to subtract one(as you have mentioned in the third paragraph).

For eg: if n = 11 then, for 2nd LSB we will have (11-0+1)/2=6 which is even. To be correct we will again have to subtract one from 6 to get 5 (which is the correct value of distance).

ThoughtNerd

4 years ago, # ^ |

Thanks a lot

kuchnahiaata

-8

Can you explain how this works,

betaphoenix

Number of swaps in LSB bit is n, for second last bit n/2 and so on...

mala07

In case you are asking for C See how each bit changes consecutively for the sequence of numbers from 0 to n , for example, consider $$$0th$$$ bit it changes(means 1 to 0 or 0 to 1) every consecutive element (see it as $$$2^0$$$) now consider the second bit it changes after $$$2^1$$$ times similarly 2nd bit after $$$2^2$$$ times... so the answer becomes $$$n+n/2+n/4$$$....till zero (each term for each bit).I hope this helps.

dv.jakhar

Bro, you are not failure, thanks for great explanation!

arthou

-10

I wrote "2*n — #bits set" but it gave WA. Pls correct the tutorial/editorial.

EDIT: Sorry @below, you missed the absurdism

codemastercpp

did you make sure that you were checking all 64 bits? For ex, in C++, cout << 2 * n — __builtin_popcountll(n);

use: __builtin_popcountll(n) and not __builtin_popcount(n).

_Anam_

please explain how did this formula come ?

anshul565

Let say for ex. we have 11010 which is equal to 10000+1000+10 hence for 10000 we have (2^(4+1))-1 for 1000 we have (2^(3+1))-1 for 10 we have (2^(1+1))-1 because for 2^k answer is (2^(k+1))-1 as explained in ED. so now we have ans as ((2^(4+1))-1)+((2^(3+1))-1)+((2^(1+1))-1) therefore ((2^(4+1))+(2^(3+1))+(2^(1+1)))-(1+1+1) and as we know this ((2^(4+1))+(2^(3+1))+(2^(1+1))) is shifting one bit to the left i.e. multiply by 2 and (1+1+1) are the number of bits where we have '1' hence ans is 2*N-number of set bits Hope this will help

tiktak2

for 10000 why (2^(4+1))-1 ?

pls, anshul565

first 10000 means 16 in decimal so for 10000 i.e 2^4 we can calc it by using what we know from ED i.e. for 2^k ans is (2^(k+1))-1 hence for 10000 we have (2^(4+1))-1 Summing these up we get that the result for n=2k is equal to 2k+1−1=2n−1.

MyAngelBakapiano

wow fast editorial

saurabhsingh1998

+34

btt does'nt seems good !

Sutaner

Is there anyone "Wrong answer on test 41" in div1.A like me ?

shubhammitt

There are many. Just use the filters in status tab.

I just use a set to count the number of adjacent topic which less than or equal to the current one, and forget to specially judge the topic with the same id with it...

spookywooky

lol I wrote in my code as comment

/* check if possible
 * adj must contain all smaller numbers, and must not contain i.
 */

But did not check for "must not contain i" ;)

dantrag

Wait, but doesn't it contradict this requirement: Each pair of blogs that are connected by a reference has to cover different topics because otherwise ? I assumed there are no edges connecting the same topic.

Thallium54

I made the same mistake and I feel very stupid now :(

UPWale

I got WA on test 41 too.

saquib2508

Did you notice that m and n can be 5e5? I can't see your code at this point, so trying to guess.

Well, for me it was Div2D, but same. I couldn't imagine a edge case here :/

anton_200

Can you help me why does checking every neighbour not give TLE? There can be many neighbours of each node right so won't it be O(n2)?

HelloWorld

Checking each neighbor means iterating through each edge twice. There are 2*m computations, that's why it won't give tle.

yebuahc_hbaruas

div 2 C had so many solutions!

1.  2*n-setbits(n).

2. authors solution.

3.  my solution which I found after an hour of analysis-all 2*n give 1 , all 4*n+1 give 2,all 8*n+3 
   give 3..and so on...

yup, I too found a pattern like 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 but wasn't able to code it during the contest.

joaom

But that pattern is wrong, the right pattern is like 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1

jac_nikola

I too got the same pattern. My solution: 82542445

can you explain your solution?

You can see that there's a pattern of 1,2,1,3,1,2,1,... Let's say n = 21. Binary representation of 21: 10101

Observations:

Number of 1's in pattern = c1 = ceil(21/2) = 11

Number of 2's in pattern = c2 = floor(21/4) = 5

Number of 3's in pattern = c3 = ceil(21/8) = 3

Number of 4's in pattern = c4 = floor(21/16) = 1

Number of 5's in pattern = c5 = ceil(21/32) = 1

We take ceil() when ith bit is 1, and floor() when ith bit is 0. (1 <= i <= |s|) Hence, answer = c1*1 + c2*2 + c3*3 + c4*4 + c5*5.

aramun_haesulla

+98

3.141592653

Please don't leave out constructforces XD

P.S. No offense to those who hate constructive algorithms. Just a joke :)

xodeChemist

recently bit storm started

Flower08

+145

shiftyblock

-92

Div 1 A was so confusing, their output makes ABSOLUTELY no sense. Why can't i just print 1 2 3 instead of 2 1 3?

Order does not matter!

mystery__coder

The desired output is in the order of number of blogs in which they were filled. Please read the question properly. 1 2 3 is not the correct output.

TheScrasse

If you print 1 2 3, the topics on the nodes adjacent to node 1 have not been set yet, so node 1 would have topic 1, and not topic 2 as desired.

It took me more time to solve A and B then C and D combined. I did not freak out initially and so was able to eventually solve C and D. Overall a great contest for me. Hopefully I will become an expert.

int128.

+86

This round tests participants' English level.

Solution C is a sequence from OEIS. $$$a(n) = a(floor(n/2))+n$$$

Cybuster

I'll leave the link with the reference for those unfamiliar with OEIS https://oeis.org/A005187

code_warrior

Will the solution given by editorial of problem B of div 2 not give tle? Since,the time complexity will be O(T*N*N) which is O(N*N*N)!

MatesV13

No, because sum over all $$$N$$$ is at most 1024!

So you can't look at it like maximum of $$$N$$$ and $$$T$$$.

+10

ok,so does it mean if x1+x2+x3+...+xn<=k => x1*x1+x2*x2+....+xn*xn<=k*k

ehnryx

x1*x1+x2*x2+....+xn*xn <= (x1+x2+x3+...+xn)^2 <= k*k

-11

Even if it were $$$T*N*N$$$, not all of the last step would reach $$$N$$$ if you check for existence and break early. Though that could be quite tight.

rahulrawat09

"It is guaranteed that the sum of n over all test cases will not exceed 1024"

nishit.sharma

Pretests for Div2 C should have been stronger.

vm.ded

Can someone please explain why this code for Div2 C fails:

Try using your binary power function, inbuilt pow is quite unreliable.

Your code works, you just have to note that

pow returns a double so you should cast it back to an integral type before doing further computations in order to preserve precision.

I made this fix an it passed:

johutha

Let me just drop that link here: https://mirror.codeforces.com/blog/entry/21844

G.O.K.U_Instinct

Same happened with me also. I used inbuilt log2() function present in c++ but got TLE on test case 8. Here is my code https://mirror.codeforces.com/contest/1362/submission/82540420

int fun(int n){

int x=1ll<<((int)log2(n));

if(n-x==0)
    return (1ll<<((int)log2(x)+1))-1;

return (1ll<<((int)log2(x)+1))-1+fun(n-x);

This was my function which calculate the answer (sorry for bad implementation) which got wrong due to inbuilt log2 function:(

2^59=576460752303423488 => (int)log2(576460752303423487)=58 but it given ans as 59 and due to this my submission got wrong.

This happened due to floating point errors caused by c++ inbuilt functions. After this i come to know that never trust these inbuilt functions:(

tejustiwari

← Rev. 5 →

use the function written below .. instead of inbuilt log2 function. P.S. " this will work only for gcc compilers "

int log(long long int x)
{
return 64- __builtin_clzll(x) -1;
}

noob_ptr

In Johnny and His Hobbies question will the given solution not give time limit as t value is also upto 1024 and we are using O(n*n) solution so it can go upto 10**9 ??

It is ok because sum of all N over all tests is small, just 1024.

Look at this comment , and also check others below, maybe it will help you.

That is the point of the statement from the Input sectoion: "It is guaranteed that the sum of n over all test cases will not exceed 1024."

swapnilr

In the Div2B editorial:

It uses the observation that if $$$k$$$ satisfies the required conditions, then for every $$$s \in S$$$ there exists such $$$t \in S$$$ ($$$t \ne s$$$) , that $$$t \oplus s=k$$$.

Can someone give proof of this?

navneet.h

if not, then k would not have satisfied the condition.

Given that $$$k$$$ is the answer, lets observe some $$$a$$$ — it will become $$$a \oplus k$$$. Since we said $$$k$$$ is answer, then $$$a \oplus k$$$ is also a number from the original list — that will become $$$a \oplus k \oplus k = a$$$ !

You can go as far as saying that numbers must come in pair!

Thank you, now I got it.

art-hack

Seems like the pretests were pretty weak in Div2C and Div1A, a lot of solutions failed in System Testing.

Will the solution given by editorial of problem B of div 2 not give tle? Since,the time complexity will be O(T*N*N) which is O(N*N*N)! Please if anyone knows ,please explain me!

In general sum of n over all test cases will not exceed N(1024) means if you put all the testcases in 1 testcase the max n is N(1024). Treat them as 1 testcase and then analyze the complexity for given constraints.

tushar_b18030

DIV2 B -> gave TLE in system testing on test case 7 https://mirror.codeforces.com/contest/1362/submission/82518343 Can someone please explain the reason

opPO

I have the same solution in 530ms. Try to use some data structure like set. My code: https://mirror.codeforces.com/contest/1362/submission/82564925

Sibin_Thomas

C was simple. I saw the pattern but couldn't implement it on time.

Sanjeev_Nu

i have found pattern like 1,2,1,3,1,2,1,3... but my pattern doesnt work can you help why?

for every power of 2, change increases by 1. 2^0 = 1, 2^1 = 2, 2^2 = 3, 2^3 = 4.

So the pattern will be 1,2,1,3,1,2,1,**4**(not 3)

I also got this pattern but could not figure out how to use this for to compute the value for n.

every odd gives 1, every mult of 2 gives 2, every mult of 4 gives 3, etc. so compute the amount of of 2s,4s, etc. by division

coder1033

This is exactly what I am doing. For each iteration I am counting the number of 1,2,3,4.. so on,

But I don't no why I am getting WA on TestCase 4. More specifically it fails for larger values of n.

from math import ceil

for _ in range(int(input())):
    n = int(input())
    i = 0
    ans = 0
    while (1<<i)<=n:
        ans+=(i+1)*ceil((n-(1<<i)+1)/(1<<(i+1)))
        i+=1
    print(ans)

I think editorials solution is complecated. Actually

First bit changes every time.
Second bit every second time.
Third bit every 4th time.
...

So we just have to collect the sum:

    int ans=0;
    for(int i=1; i<=n; i<<=1) {
        ans+=n/i;
    }
    cout<<ans<<endl;

Yeah, I have the same solution. I think it's easier to understand.

DeadlyCritic

In problem D1A, \textbf doesn't work. In fact some of LaTeX's text commands doesn't work in codeforces. (but works in polygon)

Example :

\dots \ldots \textit \texttt \textbf \t \url

dalgona

Does there exist an O(n) solution for Div2 B?

I belive so!

Idea

Problem

If anyone knows how to solve "Problem", i would really appreciated it! :)

alex223

I have the same idea, but failed 14th test when we have odd number of pairs and xor of all items = 0 :-(

Fix it and now it is the fastest solution so far https://mirror.codeforces.com/contest/1362/submission/82600948

HadyKo

your solution is still O(n^2) right?

Yeah, but only for 1 case

Cool bro, if you find how to solve that problem when it is $$$N^2$$$, I would love to hear it!

Radhe_Loves_To_Eat

MatesV13 take any element x from the set , now if such a k exists then , this x becomes x xor k, then some element must become x , so x and x xor k must exist in pairs, So let say there are 2 elements x and y = x xor k, now k = x xor y, so hence k is one of the pairwise xors of all elements in the given set . All these elements are from 1 to 1023 , which means it has at most 10 bits which are meaningful.

Let's find what is the maximum xor you can get from any 2 elements of this set. In worst case i can choose any element x from the set, and choose another element such that if the each bit is set in x , it is unset in y, and if bit is unset in x , it is set in y, So this way i will get all set bits , since 0 xor 1 is 1 and 1 xor 0 is 1 hence the maximum i can get would be 1023.

Hence i can say that if a k exists it is between 1 to 1023 , Now iterate for each k from 1 to 1024, make a new set in each iteration and compare with original set , hence this is O(n^2)

Having macro #define int long long in code for div2B Johnny and His Hobbies gave TLE at test case 7 in the system testing, after removing this macro it passed all the test cases :(. Can someone please explain

It is probably because of tight TL. XOR between two long long is slower than between ints, they have more bits, and you are not using them, so you can save time by just using int.

I know XOR should be superfast, but it's like division and modulo, they should work in O(1), but modulo is like four times slower...

Thats that, I'm guessing your passed solution had tight TL? So thats explanation, it's just slower!

passed solution is 670ms.

PRESS_F_2_PAY_RESPECT

I used #define int long long but got Accepted :p

Can u please check below codes sir[WrongAnswerOnTest1] https://mirror.codeforces.com/contest/1362/submission/82518343 https://mirror.codeforces.com/contest/1362/submission/82567052

What should I check? I think MatesV13 is right.

that's actually a really interesting case where long and int makes a big difference, gotta keep that in mind

My question is that my JAVA solution only takes up like 300ms...

I just used a hashset and compared sizes and numbers, then printed out if it was valid or not. I don't see how you could ever get a 670 ms or even a TLE on this problem.

RedDreams

div2C has easier solution.

If we write down binary representation of numbers from 0 to n in a column. We can notice a pattern:

the first bit(counting from the right) changes its value each time as we go from 0 to n

the second bit changes its value after every second number

the third bit changes its value after every fourth number ...

gand-phati-rating-giri

I dont understand, the solution discussed for B is n^2 complexity right and for 1000 tc, the complexity will be 1000*1000*1000

AceKing

It is guaranteed that the sum of n over all test cases will not exceed 1024.

Harshil_

Spoiler

I see, I got confused because generally its 1e5, and I take it for granted that we need to perform O(n), function but here we needed to do n^2

jtnydv25

+25

For a more general version of the variant you proposed in C (which, sadly is how I read the problem in the first place), there exists an answer if and only if the frequency of the maximum frequent color is $$$ \leq n$$$.

Proof : This condition is clearly necessary as for each $$$i$$$, at max one of $$$p_{2i}$$$ and $$$p_{2i+1}$$$ can have color $$$c$$$.

For sufficiency, consider the strategy where we always choose the most frequent remaining color $$$c$$$, other than the current color (the right end of the rightmost pair) and append a pair such that $$$c$$$ is its left end. We will maintain the invariant that no color has frequency more than half of total remaining length. If $$$2 m $$$ pearls were remaining and the maximum frequency was $$$ \leq m - 1$$$, the invariant holds. Else, if there were two colors $$$p$$$ and $$$q$$$ with frequency $$$m$$$, one can either append a pair $$$(p, q)$$$ if it exists and if it doesn't exist, there must be atleast one pair $$$(p, p)$$$ and one pair $$$(q, q)$$$, and we can append these pairs in one of the two orders. Clearly, the invariant holds for the remaining length of $$$2m - 4$$$. If there is a unique color with frequency $$$m$$$, and it is not equal to the last color, the invariant holds as well.

Only remaining case is when there is only one color with frequency $$$m$$$ and it is the same as the last color. We prove that this never happens. For this, go back till the right end of a pair is not equal to $$$c$$$ (say at position $$$2i$$$), or to the beginning if there doesn't exist such a pair. In front of this pair, no pair could have had left end as $$$c$$$, else we'd have frequency of $$$c$$$ > half remaining pearls at that point. But after the position $$$2i$$$, $$$c$$$ must have been the only maximum frequent color and could have been added as the left end, so a contradiction.

Now, at the end of the process, if the first and last colors are different, we are done, as we can complete a cycle. Else, say the sequence of colors is $$$p_1 = q, p_2, \ldots, p_{2n} = q$$$. Find the last $$$i$$$ such that $$$p_{2i} \neq q$$$ and $$$p_{2i + 1} \neq q$$$ and reverse the suffix after it. If there doesn't exist such an $$$i$$$, the frequency of $$$q$$$ must have been $$$\geq n + 1$$$, as $$$p_1 = p_{2n} = q$$$ and for each $$$1 \leq i \leq n - 1$$$, $$$p_{2i} = q$$$ or $$$p_{2i+1} = q$$$.

[deleted]

lohit_97

+22

Does someone mind explaining why Div.1 B works with the given solution? I still don't understand how it works!

AM_I_Learning

Setting OEIS task must be banned it took me 60-70 minutes to just realise the pattern and solving the formula with bitmasking and observations after this task I was not having much time to solve an easy DIV2 D which was purely implementation based and I suck in implementation ,really I suck.

And that compared to another person who just googled it out breaks the spirit of competition and I think codeforces is not meant for such activities . Hope problem setters will not set OEIS tasks in future. Actually those who solve or do the competitive programming for learning or as a hobby and purely for improving their problem solving skills will never google out this things but such contestants are fewer and too fewer nowadays as ratism is prevailing over the programmers and they want to get ratings either by hook or crook ,which in long run is not gonna benefit. Thus Hoping my comment provides message to all those who are googling solutions during the contest.Please be motivated and happy coding

_Crimson_

You can think of C as an "oeis" type task. To solve it, you have to cast a spell "Oh, it must be on oeis". Than you simply generate first few members of the sequence and google it

vss16996

It would have been nice if the Problem Statement of Div1 A was more clear or Explanation of the 3rd test case was available from the beginning.

Manan_shah

+45

The statement of D was very bad. It took me 15mins to understand.

mpily

in div1A/div2D since there is no restriction on blogs for which neighbours have not been written, shouldn't a case like

6 5
1 2 
1 3
1 4
1 5
1 6
5 1 1 1 1 1

be doable by writing blog 1 first and then writing the other blogs in whatever order?

The topic of a blog is always the first unused topic of all adj, so it is always restricted.

But for the first blog, since none of its neighbours have been written why should its value be restricted to 1? For example in the example i gave before writing the first blog none of 1's neighbours have any values so it can be set to any value as the described algorithm of topic allocation does not specify what value it should take

It is restricted to 1 because the statement says so.

"...and chooses the topic with the smallest number which is not covered by neighbors."

I guess you are right. Thanks.

The_Kings_Gambit

+16

To Contest Committee: I think it's time to feel sorry first for the weak pretests and further for the weak systests.

Here is a submission for D:https://mirror.codeforces.com/contest/1362/submission/82565932

And it gets AC despite failing on:

mrpandey

My submission fails the above test, yet it passed the system tests for Div2 D. A similar test case is:

2 1
1 2
2 2

_apurv_

Lol, woke up this morning and was looking at my D solution proudly, and then I noticed I had used == instead of <= in an if statement by mistake. So, here's another :

But I am really surprised, how was this not a part of the systests lol. They messed up, but still it's fine, shit happens.

wassim.omais

In Div. 1 B, if you were lazy and did not want to (for whatever reason) keep track of $$$s/p^k$$$, it is also possible to pick some prime numbers from a large set of primes (let's say, from the first $$$10^9$$$) at random. If you pick enough of them, then with high probability, $$$s = 0$$$ only when these primes divide $$$s$$$. A good bound can be proven by noting that $$$p^{10^6}$$$ can have at most $$$\approx 10^6$$$ unique prime divisors. So the chance of a comparison failing would be at most $$$10^{-3m}$$$, where $$$m$$$ is the number of primes chosen. Thus, you can do some operations modulo these primes. It was actually possible to pass in the contest by picking some large hardcoded primes, but you would risk getting hacked. But without hacking, there's not really a sure way to block these solutions.

E.g., see this solution, which got uphacked: https://mirror.codeforces.com/contest/1361/submission/82528165

vviisshhaall

Can you please explain why the greedy solution in the editorial is optimal.

clyring

+55

I suspect from the number of solves on Div1C/Div2F that many people, even in Div1 (myself included), did not know off-hand how to efficiently construct an Eulerian cycle when one exists.

Testing for existence of an Eulerian cycle is well known and easy enough to implement: A graph has an Eulerian cycle if and only if no vertex has odd degree and the vertices of positive degree form one connected component. The constructions for such a cycle I had seen, however, are not easy to directly apply to a problem of this size. The approach I came up with, which I wasn't quite able to implement within contest time, was to use (implicit) linked lists to efficiently combine cycles in the graph, and to find cycles by greedily following and removing edges not already used until there are none left.

This cycle-finding approach uses a simple invariant: At each step, either the current vertex is the start vertex and every vertex has even degree, or the start vertex differs from the current vertex and these two are exactly the vertices with odd degree. In particular, there is always another edge to follow except at the start vertex.

To create a full Euler cycle, pick an arbitrary starting vertex with positive degree, and generate some cycle from that point. Then, traverse this cycle around the graph, but whenever a vertex is reached that still has positive degree, call the basic cycle-finding method at that vertex and insert the basic cycle found at the current point in the cycle being traversed. To see that the resulting cycle includes every edge and is thus an Eulerian cycle, note that it contains every edge out of every vertex visited by the cycle. Since the set of (initially) positive-degree nodes is connected and it starts at some (initially) positive-degree vertex, this means that it visits every (initially) positive-degree vertex and hence includes every edge.

My solution, as it was ten minutes after contest end (i.e. unpolished and poorly documented but functional): 82567455

This isn't quite the approach taken by the solution in the editorial, which (at a quick glance) uses recursion to accomplish something similar using vectors. I thought about taking a similar approach, but was afraid a stack overflow might be possible on some test cases since the recursion depth might reach 250000 on malicious inputs unless special care is taken.

Svlad_Cjelli

It looks like you were right to be worried about recursion. My recursive DFS solution timed out, and it was fine after I switched it to a stack. Really annoying though...

The tutorial solution is tail-recursive, which may be the difference, although I am not sure how much the C++ compiler optimizes it.

Which function is tail-recursive in the editorial code? For some reason I'm not able to find it; it seems like just regular use of recursion to me (e.g., in function go(), there is definitely code following the recursion).

You are right — it's not tail recursive (not sure what I was thinking).

It's still a bit of a mystery why my original solution doesn't pass — after comparing my dfs with the tutorial, mine doesn't seem noticeably heavier. Maybe I just did something else wrong (infinite loop?).

EDIT: After looking over it, my dfs was simply inefficient (for high-degree vertices it went through all their neighbors multiple times). So recursion should work fine.

achintnarang10

Can anyone tell me why my solution for B is showing tle? https://mirror.codeforces.com/contest/1362/submission/82505539

flying_dog

Instead of endl try using '\n'.

sandeep_kumar_e

you have written an O(n^3) solution try to scale it down

shivaram_22

In Div2 problem A if we fix the size of max number of bits to 64 and had allowed overflow then can somebody please suggest me a solution

Axthe_E_I

Solution of C is very easy:

while(q--){
    long long int a,c=0; cin>>a;

    while(a){
        c=c+a;
        a=a/2;
    }
    cout<<c<<endl;
}

____ALPHA____

Thank u. Can u explain how it is working? I mean, math behind it ?

Let any number(for e,g. 10 ), then we have to find sum of bit differences from 0 to 10, but instead of comparing 0-->1, 1-->2, 2-->3,... we will take different approach, first let write the numbers in bit form,

0 -  0 0 0 0
1 -  0 0 0 1
2 -  0 0 1 0
3 -  0 0 1 1
4 -  0 1 0 0
5 -  0 1 0 1
6 -  0 1 1 0
7 -  0 1 1 1
8 -  1 0 0 0
9 -  1 0 0 1
10-  1 0 1 0

Consider it a matrix of [11][4], so instead of going top to down, we will go comparing right to left (cause we only need sum of all differences so it won't matter sum will be same).

LOGIC: In the matrix if we go from right to left, we see the last column of the matrix every bit differs by it's neighbouring bit {0,1,0,1,....}, so, sum += 11; Next, second last column every 2nd bit differs by it's neighbouring bit {0,0,1,1,0,0....} so, sum += 11/2; } similarly if we do same till the first column, we get the sequence,

while(n){
     sum+=n;
     n=n/2

gegewepe

nice explanation

TleTerror

language was very tricky in this round...overall good contest...thanks

darklight13

I think the time limit for Div1 B could have been relaxed to 3/4s as the modulo operation is slow. Many solutions failed even if they had the desired asymptotic complexity.

chuka231

+19

Div2D/Div1A — Johnny and Contribution The hardest readforces statement ever seen.

jha_saurav

I feel there is some issue in question A, i used a simple approach of dividing the maximum of them by minimum and then that dividend must be in powers of 2 and then i applied log to the the power value.

After this i simply from 3 two'2 to get 8 2 two's to get 4 and 1 two to get two, and it's running for the test cases in the problem.

Please help out.

link to the solution : https://mirror.codeforces.com/contest/1362/submission/82512703

Thank you

Since in this testcase the answer is 20, the numbers are fairly huge, like 1 and 1<<60

Not sure if this is handled correct in python.

aKILLL

I have done the same in C++ getting WA in the same test case don't know where the error is

For Div2E/Div1B-Johnny and Grandmaster, why does the strategy given in the editorial works?

Karan123

So, i understood the editorial in the following way:

First, if p>1, then we sort the array(that contains powers of p) in non-increasing order. Now, we have a variable $$$s$$$ that will denote the difference between the two partitions. Initially, s = 0. Now, If s = 0, we add the current element to sum.

If s>0 then we subtract the current element from s. The result will be non-negative. This is because: If current element is $$$p^x$$$ and s>0, then let $$$s = p^{a} - p ^{b} -p^{c} + p^{d}$$$ (The operation among $$$p^i$$$'s can be both addition or subtraction), where a,b,c,d are non increasing sequence. Hence, $$$s = p^d*(p^{a-d} - p ^{b-d} -p^{c-d} + 1)$$$. The second factor can only be positive since s>0 (If second factor is 0 then s = 0). Now if second factor is 1, then s = $$$p^d$$$ and $$$p^x$$$ is less than or equal to $$$p^d$$$. So after subtracting it from s, s can be 0 or positive. If second factor is greater than 1, then $$$s>p^d$$$ the subtraction will yield a positive s.

Note that till now we are not doing modular addition or modular subtraction. Now, at any element, if $$$\frac{s}{p^x}$$$>n, implies $$$s > n*p^{k_i} $$$, implies $$$s$$$ is so large that even if the rest of the elements are equal to current element (which is largest possible value for them), s will not become 0. (I think here n is the length of array that is not processed yet). So we know that all the other elements belong to one partition. And after this moment, we can do modular subtraction for the rest of the elements.

NeverGonnaGiveYouUp_

Do you know how to check if s > n * p^k_i ? I can't understand that part of the editorial. Thanks

Kiona0405

Thank you great posting. I have one question about modulo operation. Why can't we take module before $$$\frac{s}{p^x} > N$$$ is satisfied?

Because we need the s value and not s%MOD to take the decision whether the current element goes to first or second partition. Let's say s becomes MOD(so s%MOD becomes zero), then we will add the current element instead of subtracting it. Having s = 0, would mean that both the partition have equal sum and we add the current element because we can keep the current element in any partition. But having s%MOD=0 means that one partition might have MOD problems more and in that case optimal step is to subtract the current element.

Also, if we consider s%MOD in the condition $$$\frac{s}{p^x}>N$$$, then we want s to be a very high value to satisfy but s%MOD will never exceed MOD, so if we take s%MOD before the condition fulfills, we might never achieve the condition.

I could finally understand!!

lucasxia01

You can just calculate the sum modulo the whole time. To check whether to add or subtract, you can use $$$s/p^i$$$ instead, since if $$$s = 0$$$, so does $$$s/p^i$$$.

Hey, you described the solution very well. But it does not prove why is this optimal solution. https://mirror.codeforces.com/blog/entry/78355?#comment-636627 . This comment gives the proof.

Confused_Unga_Bunga

Not solving C cost me Another Rating Drop

taay_hannan

kripya yaha par na royein

jainvaidic

1362C — Johnny and Another Rating Drop

This was my solution(82558750) but last test case (test case 8) input 2**59 — 1 it failed

log2(2**59 -1) = 59 was evaluated by my code and was giving answer of 2**59 can anyone help what should I do for that

_sv_

In C++ log2 uses floting points with precision up to 1 decimal degit and hence gives the wrong answer. Changing log2() to log2l() helped me to get AC.

other way is to use bits taking & with (1ll<<i)

aneee

Can someone pinpoint what the error is? Problem A (Div. 2) Submission Expected -1, But printed 20.

You are using floating point arithmetic, your values are casted between long long and double. A double uses less than 64 bits for precision (I think 56, but not sure). So at some point you simply get presission loss and one of your log2() or ceil() call returns not the expected value.

goalfix

82561759 Hey, beginner here. Would you mind taking a look at my submission too? I failed at the same test case but I used a while loop instead... is this cause for error similar and how do I fix this in python.

Thanks in advance!

Edit: I saw your reply on python on an earlier thread. My bad!

It seems that it is complecated in python to say how much bit an int has. stackoverflow

Since the testcase is the one where $$$a==b«{60}$$$ I think that is the problem.

Thanks for your reply! So is there a better way to check if a number is an integer? I didn’t want to keep multiplying 2 in a while loop, because some easy math will directly give the power of 2 with which we should multiply.

The problem with floats is that they are silently truncated. So there is actually no way to check if a double is really an int. All you can do is checking if it looks like an int. If it does it maybe is an int, or the part which is different from an int is so small that it was truncated.

This is error prone. Basically everytime you think you need to check if a floating point variable is an integer your implementation is simply wrong.

True.

So maybe after computing x = floor(log2(b/a)), I can check if b == a*pow(2,x) ? That should take care of the issue I guess.

Thanks again, learned something new!

destroyer18

nice round fast editorial language of div2D/div1A was too heavy..took me around half an hour to understand....but could not solve during the contest

Okrut

Auto comment: topic has been updated by Okrut (previous revision, new revision, compare).

rahuliitkgp

Another solution for C

Let the position of rightmost zero in nth number be ith position then we know in next number (n+1 th number) all the bits from 0 to ith position will be flipped and only these bits will contribute to our answer so the contribution of any nth number to total answer depends on the location of rightmost zero bit in n-1th number. so we want to find how many numbers are there which have all bits 1 up to i-1th bit and has a zero on the ith bit. It turns out that the number we are looking for has form k*(2^(i+1)) + (2^(i)) — 1. (i = 0, 1, 2 ...) equate it to n for each i find k(take ceil) multiply it by corresponding i and sum them. the total sum will be the final answer

dm99

For Div2 Problem E, can someone please explain me why a simple approach like this does not work?

My guess is that it is because of modulo operation. Can someone explain a little about how it is being handled in the author's solution?

_Enigma__

Because the (ans==0) condition may be true even when the absolute difference is not 0. Your solution will change ans to curval (i.e. add curval to it) whereas it may be required to be subtracted.

This solution can actually pass, if you check the answer modulo multiple primes rather than just one. But it is not as safe, and if you hardcode them, it’s possible to hack.

shubhampoddar786

Can you explain with an example?

If we take p=31623 and then the values of ki's are: 2, 0, 0, 0, 0,.......(around 15000 zeroes), then there will be a case when difference becomes 10^9+7 but since we are keeping it modulo, it will be considered as 0 in the code.

And the next 1 would be added making ans=1 at that step while correct value should be 10^9+6 at that step.

Alapan

-15

The writers dont know how to write questions. The Div1 A was kinda encrypted!

talibmohd

In Div2 E/ Div1 B, How do I prove that for the optimal solution, the sum of the partition containing the highest power will always be more than the sum of the other partition?

Aoi_Hill

3 years ago, # ^ |

Correct me if wrong but I simply thought in base 'p' representation of number , so like binary we are focusing on comparing the MSB.

sokokaleb

+13

Div1 C was an enjoyable problem, even though I couldn't get it done within the contest. Thanks for making this particular problem! :)

jerryliuhkg

I don't really understand the editorial of Div. 2 QE, can anyone explain? Thanks!

Dragon_in_Bed

Me neither, I have no idea what the tutotial is saying. Can someone help explain?

Sort the $$${k}$$$ array in non decreasing order. Now start from the right most index say $$${n}$$$
Suppose if $$$\sum\limits_{i=1}^{n - 1}{p^{k_{i}}} < {p^{k_{n}}}$$$ then we put $$${p^{k_{n}}}$$$ in one set and the remaining powers in other set and compute the powers and their difference.
If $$$\sum\limits_{i=1}^{n - 1}{p^{k_{i}}} >= {p^{k_{n}}}$$$ then $$$\exists j : \sum\limits_{i=j}^{n - 1}{p^{k_{i}}} = {p^{k_{n}}}$$$. Then we can put $$${p^{k_{n}}}$$$ in one set and $$$\sum\limits_{i=j}^{n - 1}{p^{k_{i}}}$$$ in other set to make their absolute difference 0. Now go to index $$${j - 1}$$$ and repeat the process.
This can be implemented by using a stack and processing the $$${k_{i}}$$$'s from the right most index to the left after sorting in non decreasing order. When you see $$${p}$$$ $$${k_{i}}$$$'s in stack remove them and convert it to one $$${k_{i}}$$$ + 1 and proceed further.
Here is my submission for reference 82581557

adhvana

How to prove that there exists such j ? UPD — Convinced myself by thinking in terms of base p.

Can you write how you are convinced in more detail here

I will try — $$${p^{k_{n}}}$$$ = $$${(00..010...)}_{p}$$$ where on-bit is position $$${k}_{n}$$$. Assume $$${k_{n-1}}$$$ is less than $$$k_{n}$$$ otherewise take them in different sets and proceed. Let $$$j$$$ be the largest index such that adding $$$p^{k_{j}}$$$ will result in sum $$$greater$$$ $$$than$$$ $$$or$$$ $$$equal$$$ $$$to$$$ $$${p^{k_{n}}}$$$. Before adding $$$p^{k_{j}}$$$ state will look like — $$${(00...(p-1)(p-1)(p-1)(p-1)...(p-1)00..00)}_{p}$$$. Earlier digits are not set because we are proceeding in non-increasing powers.

xyz_

I tried to simplify the implementation and add comment- 82689935

Noumi

Can anyone please tell me the proof for div2 E

I mean the proof for it's correctness

Sorry I'm not able to frame words to explain its correctness.

we_ak_covid_19

Thanks for the solutions! But I wonder why my solution didn't work last night. I checked how many bits the greater one has to shift in order to get the smaller one. I greedily checked how many 3's, 2's and 1's I can shift. However, it didn't work. Can you help me out? Thanks!

Riyuzak251097

My solution for DIV 2 E seems to give Time limit exceeded on test case 7 .. Can someone point out the issue in my code.. it will be very helpful My Submission

cote

Could anyone please help me debug my code for problem div2A: 82590156. I get wrong answer on test case 2. My idea is to check q = b/a. if q is a power of two then greedily count the steps, if it is not then print -1.

laksh1729

Even I was using the same logic. It seems some implementations with float numbers might fail due to precision errors. Its even mention in editorial div 2 C -- educational round 88.

Thanks, I have found out which test my code fail.

1 576460752303423483

1 2^59-5

The log2 function cannot spot the difference. Maybe I will change it to some bitwise operation.

You know the log2() function comes in severeal flawors. If you use the one working with long double instead of double it would work better.

But of course, using integer arithmetic is the better solution. I just wanted to point out that it is not the log2() function that causes the error, but the used data type.

gk24015

-14

can someone help me spot the error div 2 problem B round 647?

include

KnightKnight

I think looping it till 1024 will make it work

silent_sky

Can someone please prove the observation 2 in the editorial of Div 1A/2D?

johnny has to color from 1 to target-1 because if any of those numbers are missing then he'll color a number lower than target

harshitsoora

I have been stuck in question C like from last night.

It is showing wrong at test case four, I don't know why it is passing all other n <= 10^18

Can somebody please please help me out. Thanks in advance!!

My Solution

I was trying to do it in O(bit length of n)

Solution Explanation: Old number in (0,1,2,.....n): difference = 1 like between (1,2) or (3,4) Multiple of 2 : difference = 2 like in (2,3) Multiple of 4 : difference = 3 like in (4,5) or (12,13) Multiple of 8 : difference = 4 like in (8,9) or (24,25) Multiple of 16 : difference = 5 like in (16,17) and so on...

So I first find these multiple and then multiply with respective differences

Someone Please help, I don't understand why it not passing Test Case 4!!

twos[i] = n/pow(2,i);

I assume there are huge values of n where you do net get the expected result. Just do not use pow here, instead do 1LL<<i.

And there is no need to go up to 66, 63 is enough.

Thanks Man,

You were a day saver for me ;)

Its working great now!!

swapnil07

Can someone please tell the mistake in the solution of my Div1B solution 82594745. If I change LIM = 1<<12 to LIM = n-i, it becomes AC 82595479.
The logic is simple, I keep on subtracting the i-th number until the current value turns 0 in the reverse sorted array. If the value is 0, I make i-th number the current difference.

Found the mistake, wrote 1<<12 instead of 1e12.

_undef

could someone explain why this solution fails at A? https://mirror.codeforces.com/contest/1362/submission/82596448

try considering that case when u can divide by 4 also ! n%8 will not suffice for the greedy solution

cjj490168650

Div1C

Can anyone help me with this submission? 82597192

I don't know why it TLE.

log2 of the number 576460752303423487 is 59.0 while 2^59 is 576460752303423488 (one more)

I can't understand this unusual behaviour.

Many solutions of DIV2C got TLE because of this.

I got my answer. log() operation uses floating point numbers and offers precision up to one 1 decimal point.

hence log2 of number 576460752303423487 gives 59 instead of 58.9999999999999999975.

nweeks

Hello, why does the greedy algorithm for d1B work ? I tried proving it for around an hour during contest but I was far from being confident it was true. I submitted it after seeing so many people solving it. But I can't come up with a satisfying argument for this.

+37

By induction: The base case is trivial, as the greedy strategy is clearly optimal when n=0 or n=1. The recursive case relies on a parity argument. Suppose without loss of generality that $$$\{p^{k_i}\}_{i=1}^n$$$ is already sorted descending and we have already assigned the first $$$j$$$ terms in a way that minimizes the imbalance of only these first $$$j$$$ terms. Then, the magnitude of this optimal imbalance is clearly a multiple of $$$p^{k_{j+1}}$$$. Now, consider two separate cases: Either this optimal imbalance is positive, or it is zero.

If it is positive, then it is at least $$$p^{k_{j+1}}$$$, since it is a multiple of that amount, so adding $$$p^{k_{j+1}}$$$ to the smaller side of some optimal configuration reduces the imbalance by $$$p^{k_{j+1}}$$$. But the (reverse) triangle inequality tells us that there is no way any configuration's imbalance can be reduced by more than $$$p^{k_{j+1}}$$$ by adding that term to either week, so reducing an optimal configuration by this amount cannot be improved upon.

The case of zero imbalance is more subtle. Since switching any $$$p^{k_i}$$$ between the two weeks results in a net change of $$$2p^{k_i}$$$ and zero is an achievable imbalance, every possible assignment for the first $$$j$$$ terms will result not only in a multiple of $$$p^{k_{j+1}}$$$, but a multiple of $$$2 \cdot p^{k_{j+1}}$$$. Thus, applying the (reverse) triangle inequality again, adding $$$p^{k_{j+1}}$$$ to either side of a suboptimal configuration cannot achieve an imbalance of less than $$$2 \cdot p^{k_{j+1}} - p^{k_{j+1}} = p^{k_{j+1}}$$$. But this is obviously also achieved by adding $$$p^{k_{j+1}}$$$ to either side of an optimal configuration, so the greedy strategy again cannot be improved upon in this case.

I think this solution needs and assumes that the problem follows Optimal Substructure Property. How can we prove that this problem follows Optimal Substructure Property?

This actually is a proof that the problem has the kind of optimal substructure necessary for the greedy algorithm to work. Specifically, the recursive case of the induction argument says that if the greedy algorithm produces an optimal (sub-problem) result after $$$j$$$ steps for some $$$j < n$$$, then the greedy algorithm also produces an optimal result after $$$j+1$$$ steps.

It's easy to see that the greedy algorithm doesn't have any opportunity to make a mistake after $$$0$$$ or $$$1$$$ steps. (This is the base case I started with.) Then, applying the argument above with $$$j = 1$$$, the greedy algorithm must produce an optimal result after $$$j+1 = 2$$$ steps. But then the argument above can be applied again with $$$j=2$$$ to show that the greedy algorithm produces an optimal result after $$$j+1=3$$$ steps. And then with $$$j=3$$$, $$$j=4$$$, all the way up to $$$j = n-1$$$ to prove that the greedy algorithm produces the optimal (correct) result after $$$j+1 = n$$$ steps. This method of proof is called mathematical induction. Does that clear things up?

Got it. Thanks!

Thank you for the proof ! I really had the bad representation for this problem, it is really clear now.

parveen1981

Can you please explain this line?

But the (reverse) triangle inequality tells us that there is no way any configuration's imbalance can be reduced by more than pkj+1 by adding that term to either week, so reducing an optimal configuration by this amount cannot be improved upon.

shankamboj

I am stuck in D . Can anyone Help? Here is my Blog https://mirror.codeforces.com/blog/entry/78387

you code could have been more readable if you want someone to debug it

te_amo

Div2 problem D really tested participant's patience, concentration, and English. Really interesting set of problems. Thanks!

TheAct1on

Could you help me please on task A. Why does the checker give -1 in this test?

576460752303423483 1

Img solve

the question states divide only when it is divisible by 8,4,2 the first number you have chosen 576460752303423483 is not divisible by 8,4 or 2 So the answer is not possible

anupamshah_

The time limit of Div2-D should have been 1.5 seconds, in order to prevent several brute force solutions.

the only solution was a brute force solution wdym, and it ran in 1.3s in java so you'd be screwing over slower languages like python by lowering it significantly

xygzy

Can someone help me understand why my solution for Div2 A using log2() didn't work (WA)?

My solution: 82539547

I think in the last step you are taking ans as r/3 and then taking count++ for r%3!=0 You haven't checked for r%2 !!! I think You missed that ! You have 8 , 4, 2 and you are just assuming your division from 2 and 8

Assume b==1 and a==(1<<60)+1

Your call to log2 will cast a to double which will make the +1 go away because of precission loss. You could try to cast a explicit to long double so that the log2(long double) variant of that function is called.

Or better, do not use floating point arithmetic for this problem at all.

expected there would be meme with each problem on tutorial

Yudhishthira

For DIV2 E, let mod=1000000007, if(difference%mod==0) does that necessarily mean than difference is zero because mod is prime??

euler_3

No. (diff%mod==0) doesn't always mean (diff==0).

For simplicity, take mod = 7 (prime) and the following testcase:

n = 3, p = 2, k = [3,0,0]

Hope, you got it :)

Lithosphere

"Observation 2: If for a vertex u with color c there exist a color c′<c such that u has no edge to any vertex with color c′ then the answer is −1."

Can someone please explain this to me. Its not obvious to me at all.

shahincsejnu

suppose, we are at node 4 and its color is 5(according to given/desired sequence/coloring), and its neighbor nodes have color 2,3,4. And there is another node lets call it x and it has color 1. But node x is not referencing node 4, then according to the condition we have to color this node 4 with the minimum number of color which has not taken by its neighbor (which is 1 for node 4 in this case), but according to given/desired sequence it has color 5.

So, contradiction happens. That is why always a node has to be connected to all of colors node (at least one for each color) less than his color. I hope you understood :)

utsavparakhiya

Easiest solution for the Div 2 C

solution link : https://mirror.codeforces.com/contest/1362/submission/82613654

nFlamel

Can you explain the logic?

Every bit will be changing after pow(2,i) numbers where i is 0 based index from LSB.

You can writer binary upto certain numbers and check yourself..

Lelouch.Lamperouge

For anybody who didn't implement the solution for DIV 2D thinking of it as a brute-force solution(like the process of going to each node and visiting all of its adjacent nodes ) here's the proof for time complexity

PROOF

LESSON LEARNT

It might be a very obvious thing to observe for many but I was not able to prove it in time during the contest so just thought of writing it so others like me can get it too. Please correct me if you find anything wrong or unclear.

drastogi21

Well, this is what dfs is about.

Yeah, you are right but you won't visit a node that is visited once again in dfs hence that is $$$O(n+m)$$$.

_chandler_

Lesson learnt should be :

Analyze the complexity of your solution more carefully.

If you keep submitting brute force solutions everytime, it would unnecessarily waste your time in contest, which could instead be utilized to think of an actual solution!

Yeah you are right thanks, edited

may_21

My solution on Div 2-D is constantly failing on test case 6 and outputs -1. Can anyone suggest the test case where it fails

Well I don't think that you actually need to do a cumbersome dfs to do this problem you can check my commented solution if you want

A neater solution to problem A using c++ built-ins

#include<bits/stdc++.h>
using namespace std;
typedef long long ll ;
void solve(){
  ll a,b ;
  cin >> a >> b ;
  ll ans =0 ;
  ll a1= __builtin_ctzll(a) ;// calculation of trailing zeroes in a 
  ll b1= __builtin_ctzll(b) ;
  ll A =a,B=b ;
  A=A>>a1 ;
  B=B>>b1 ;
  if(A^B){
    cout << -1 << endl ;
    return ;
  }
  ll val =abs(a1-b1) ;
  ll rem = val%3 ;
  ans = val/3+(rem+1)/2 ;
  cout << ans << endl ;
}
int main() {
	int t;cin >>t;
	while(t--)solve() ;
	return 0;
}

MrRob0t

Does the standard c++ log() function give erroneous results when used for higher numbers? For Div 2A, I first found b/a and if it is a power of 2, then I simply returned

ceil(log(b/a) / log(8) )

This gave error on the case 1 8589934592 Answer should be 11, i.e. log(8589934592)/log(8) But my code returns 12, whereas it works completely fine for the case 1 64

Can anyone please explain Div.2E/Div1.B, I couldn't get it from editorial. Thanks in advance.

samarthbtp1234

DIV 2->C my soln is working fine for smaller numbers but its giving WA for test case like 1 576460752303423487 can anyone help here is the link: https://mirror.codeforces.com/contest/1362/submission/82621109

Am I the only one who solved A with BFS?

idk321

Can someone prove for Div2E/Div1B why is the optimal sum never negative? I can somewhat see that you can for every negative sum instead find the positive sum with the same absolute value, but I can not prove it in my head at the moment.

https://mirror.codeforces.com/blog/entry/78355?#comment-636659

I understand why substracting an element will always, if the sum is positive, yield at least 0 as the sum. I am not completely sure why it wouldn't be in some case better to go in the negative sum from the 0 sum.

In other words, using the technique in the editorial the subset with the largest power has always the larger or at least equal sum compared to the other subset. Not completely sure at the moment how to really prove it, though it does seem a bit intuitive.

Let's assume that when we add a current element to sum, we add that element to first partition and when we suubtract the current element we add the element to the second partition. So, when the sum s is 0, we interprete that the two partitions have equal problems,(s denotes the difference of problems between the two partitions) so the current problem can be kept in any of the partition, Let's say we subtract the element when the sum is 0, mean we add that element to second partition and then when our sum is negative, the optimal way is to add the elements to first partition till sum reaches zero.

In conclusion, this way is exactly the opposite of what is written in editorial. If s = 0=> subtract the current element. Add the current element till sum is negative.

dendrite5460

i have a doubt Karan123 . Why does the editorial subtracts largest exponent(k[i])with 1e6+10 in the start instead of adding p^k[i] to the s ,as s is initially zero ? And why does the editorial does modular exponentiation before checking the condition s>n*p^k?

Yeah, actually I am also confused about the implementation of the idea. Actually, I came up with some other idea for implementation. First we sort the array in descending order, then according to the logic of the editorial, what we want to do is whenever same elements are present in the two partitions, we want to add the current element in first partition. Then as long as the elements in second partition are less, we keep on adding the elements in second partition. It is confirmed that as long as we know when the elements in the two partitions are equal, we can keep on adding the elements in second partition and also this process will not make number of elements in second partition more than the first partition without making the number of elements equal to each other.

Now, first I kept a variable p1 that denotes that the number of elements in first partition is $$$p^{p1}$$$. And I add the value of $$$p^{p1}$$$ to the current sum. Now, I will keep on adding all the other elements to second partition as long as the number of elements in two partitions are not equal. Now, I have created a map of int-->int where map[i] will say how many $$$p^i$$$ are there in second partition. If I have current element as $$$p^x$$$, then I will increase map[x] by one. If map[x] becomes p, that means my second partition contains p $$$p^x$$$, that means it has one $$$p^{x+1}$$$. Hence, I can make map[x]=0 and increase map[x+1] by one. Now, if it had been that before this element, there were p-1 $$$p^{x+1}$$$, then it will have p $$$p^{x+1}$$$, Hence one more $$$p^{x+2}$$$. Recursively I ensure that at a given time my map has all map[i] less than p. Now, as I move through the elements if I have map[p1] = 1, that means that the second partition now has $$$p^p1$$$ elements which means both the partitions have same elements.Whenever map[p1] is 1, I have concluded that there are total $$$p^p1$$$ elements in the second partition, since the number of elements in second partition never jumps over the number of elements in the first partition. At this moment, I conclude that the current sum is 0, and I again add the next $$$p^x$$$ in the first partition. I have implemented this idea here:

https://mirror.codeforces.com/contest/1362/submission/82699821

(Actually prior to writing this comment I was stuck at this problem, but while writing this comment I found the bug and then got AC :) )

In this idea, the catch is that the number of elements in the partition is equal when map[p1] is 1 and not when s = 0.

ajayNegi

I looked at your accepted submission and this one too and saw that in the accepted code you've used m[p1]==1 for checking if the difference has become 0 or not. Whereas in the Wrong Ans submission(the one I've linked), you used curr_sum==0 to check if the difference has become 0(this is exactly what I earlier was doing and hence getting WA). Can you tell the reason why the curr_sum == 0 not working even when we know that it is 0?

So, as I said in the earlier comment that if we are doing modular addition and multiplication, we cannot determine whether the first and second partition has equal elements using curr_sum. This is because if the difference between the two partition is $$$10^9+7$$$, then also curr_sum = 0 and this does not mean that the two partitions have same elements. Now, to keep track of the number of elements of the second partition, note that we can always express the number of elements in the second partition as $$$\sum\limits_{i}p^i$$$. Note that if $$$p^{i}$$$ occurs p times then it is equivalent to 1 $$$p^{i+1}$$$. To keep track of number of elements in the second partition use this property along with the fact that the array is in descending order and $$$\sum\limits_{i=0}^np^i < p^{i+1}$$$. Let's say you have array:[5 3 3 3 3]. Now p1 = 5 and next elements are 3 3 3 3 and p = 2. Now you will keep $$$p^5$$$ in first partition. Next since m[p1] will be zero (No $$$p^{p1}$$$ in the second partition), then keep two $$$p^3$$$ in 2nd partition. This will form 1 $$$p^4$$$. Then again keep 2 $$$p^3$$$s in the second partition, this will also form 1 $$$p^4$$$. We have total 2 $$$p^4$$$ and this will form 1 $$$p^5$$$. Hence m[p1]==1 and number of elements are same in both the partition.

Got it. Much grateful for that.

valdaarhun

Here's my submission for problem D. I am getting TLE on test 9 even though the idea seems to be the same as what many other people have implemented.

Here's another submission of mine which got accepted. It is different from the above submission in a few subtle ways such as using vector< bool > instead off bool [] or vector< int > instead of int [].

Any thoughts on why this difference exists? How do I make my implementation more efficient?

Never mind. I have got it. The problem is with my fill() statement.

gouhar

In Div2, D
For test case
5 3
1 3
2 3
4 2
1 2 3 1 4
How answer is -1.
And why output : 1 4 2 3 5 is wrong.
As node 5 is not connected to any other node, so it can be colored with color 4.

PatronSaint

As you have to assign min topic which is not covered by neighbors of the current blog. Since blog 5 has no neighbors min topic that can be assigned is 1, but the topic required to be assigned to blog 5 is 4, hence the answer is -1. I think this is the case. I am currently in the process of upsolving this question myself.

Kognition

I have a (sort of) alternate solution to Div2 F/Div1 C: The first few observations are the same regarding binary searching on the beauty and noting that for beauty b numbers can be replaced with last b bits and you can only link equal numbers, but I didn't think to make a new graph and use Euler cycle. I first check the frequency of each b bit number, to make sure each is divisible by 2. Then, I just linked necklace parts with each other arbitrarily, which will make a bunch of disjoint cycles. We notice that you can merge cycles iff they share a b bit number. So, we continuously merge cycles until we can no longer do so. If there is one cycle, then we are done, otherwise it is not possible (why?).

It is actually a huge pain in the ass to do this in linear (times inverse ackermann) time per beauty check and takes a LOT of extra memory to keep track of tons of indices, but still nice to know that you can solve this problem while forgetting about Euler cycle. I also had to do some constant factor optimizations to get it under the TL, i.e. had to be careful about using map vs unordered_map, unordered_map vs vector when possible, etc. If you want to test your implementation skills this is probably good practice.

The keen reader will notice that this solution is actually just a gross unpacking of the proof that all even degrees implies existence of Euler cycle, i.e. just make a cycle and then stitch it together with another cycle and repeat. Perhaps I should have been tipped off by the "divisible by 2" part.

decoder__

Can anyone help why my code is failing for Div2 A https://mirror.codeforces.com/contest/1362/submission/82589559

you are swapping them and that's where you went wrong. take them as it is and apply simple approach.

How does swapping make a difference? Could you please list a test case where it might fail ?

ok.so, at first look it seemed that the problem is with swapping but this is the case where your code failed:

expected -1 found 20

Thanks for the test case but I don't understand why log2(576460752303423483) is giving 59 even though the number 576460752303423483 is an odd one.

log2(576460752303423483) gives 59 because it actually returns 58.99... which gets rounded off to 59. When using such functions like this one you need to be careful with floating point precision. I got a wrong answer verdict too when using log2(). It's better to simply have a loop in which you do n /= 2 and count++ as long as n % 2 == 0.

Here's my submission with the loop -> 82602732.

sbhm_prksh

11 months ago, # ^ |

Thanks buddy

Shashank_Chugh

Can someone please explain how to approach Div2 E / Div1 B ?

Yeah, you can have a look at the simpler solution, 82595479.
We traverse the numbers in decreasing order, if the current value of answer is not equal to 0, subtract this number from the answer, else set this number as the answer.
It is definite that the i-th number cannot make the answer negative after we subtract it from the answer since p^k can always be represented as sum of lower powers, so the number will always become 0 if it has to become negative. :)

if the current value of answer is not equal to 0, subtract this number from the answer, else set this number as the answer.

I am not able to understand intuition behind this part ?

Assume p = 3 and array is 5 4 4 3 3 3 3 2 2.
Answer at each step:
Step 1. 3^5
Step 2. 3^5 — 3^4
Step 3. 3^5 — 2 * 3^4
Step 4. 3^5 — 2*3^4 — 3^3
Step 5. 3^5 — 2*3^4 — 2*3^3
Step 6. 3^5 — 2*3^4 — 3*3^3 = 3^5 — 3*3^4 = 3^5-3^5 = 0
Step 7. 3^3
Step 8. 3^3 — 3^2
Step 9. 3^3 — 2*3^2 = 3^2 = 9.

At each step, we try to reduce the current value of answer. There is no possibility of a minimum answer than this. It wouldn't have been the case if the number at i^th position was not divisible by all the numbers ahead of it. In this particular case, all the numbers can either make the answer 0 or can reduce the value of answer. We also note the answer is always divisible by all the numbers ahead of it.

bitcoin

Still don't understand why this greedy solution gets us the optimal answer

At step 1, we are sure that the answer is p^k₁. Now at second step, the number can be either equal to first number, or be equal to p^k₁-x. Now, if we add it into the same group, the chances that we will reach zero will decrease. Moreover, we observe that if there is an optimal solution by adding it to same group, then there will be definitely an optimal solution by adding it to another group.
Example: p = 3, and are = [5 5 4 4 4 4 4 3 3 3], then the optimal solutions can be [5 5], [4 4 4 4 4 3 3 3] or [5 4 4 4], [5 4 4 3 3 3]. Now if there were one less 4 in the array, the the answer would have jumped to 3⁴ whereas it could have been 3³.
At each step the answer is either 0 or the answer is a multiple of p^k_i. Therefore we must subtract it, otherwise we will be decreasing our chances of reaching the minimum value.

Thanks , you have written very nice explanation

Apology for the delay. It almost slipped off my mind. Here is the link with total explanation. 84161599

deeper-red

For Prob Div2B if n is odd the answer will only be k=0, thus no solution possible for K>1; Since a1(xor)x=a2 where a1 and a2 be any 2 numbers not arr[0] && arr[1] then x=a2(xor)a1 and thus a2(xor)x=a2(xor)(a2(xor)a1)=a1; Similarly for other pairs also. hence they occur in pairs and if n is odd we will have one pair a1(xor)x=a1 then x=0;

mstallion

My solution for div2 D is failing on test 5. If someone could help to find the mistake or a test case at which it will fail, then it would be appreciated. Here is the link to my solution: https://mirror.codeforces.com/contest/1362/submission/82657443

P___

What a terrible statement of D1C. I did not know it was that terrible until I read the editorial. When I read it I couldn't understand why can we connect only vertices with the same last bits. Then I re-read the statement 100 times more and I discovered that it should divide the xor, not be included in it...

_LRJ_

Good contest,thanks!

LRJf

I agree!But the editorial can be more specific.

Overmout

In div2 D, how to calculate the time Complexity, I can't understand it.

for every node, check every connection for the conditions.

there are only M connections, thus you have to make o(n+m) for the checking validity stage.

then, sort (nlogn) and greedy fill from bottom up. idk how u do it in o(n) but nlogn is sort.

nlogn is a bigger factor than n+m so only nlogn is included in compleixty

Thank you, so the overal complexity of that 3 Nested for loop in editorial is only o(m)?

o(n+m), because o(n) for each node and o(m) for each connection (because theres only a maximum of 'm' connections for the whole thing so some nodes u get long strings of checking and some nodes u get a little bit of checking)

That helped a lot ！thank you

Can you help me with this solution https://mirror.codeforces.com/contest/1362/submission/82657443 for div2d. Its failing on test 5.

im not too familiar w/ c++ so forgive me if i'm wrong but i'm not sure if

for(auto a : mp[node]) { if(t[a]==ref) flag = true; if(t[a]==ref+1) flag1 = false; }

is correct. if (t[a] == ref) flag = true means that if another blog has the reference the flag is true (this flag should automatcally falsify since that means this node can't be the target value anymore) and t[a] == ref+1 falsifying the node isn't correct either because it doesn't check for nodes below itself (you don't really care about nodes above taget, only below and equal to it).

remember, two conditions:

1) all connected blogs must have a set of numbers such that they are below target number

2) it must not have a blog connected such that it is equal to target

Thanks for replying. But I think you did'nt understood my solution. ref is target-1. but, I understood my mistake. thanks anyway

adhirajbist

When I solved Div.2 D using BFS I got TLE, but when I used DFS with same logic it got accepted.

BFS submission

DFS submission

Can someone please explain to me why this happened?

Consider a graph which is nearly connected(as TC 8 on which you got tle). As at each node you are checking all it's adjacent nodes assume at each node you got 1000 nearby nodes.So O(1000*n) time complexity.

since the maximum number of edges is 500000, won't the complexity for DFS and BFS both be O(V+E)?

I haven't seen your dfs code but in your bfs code for each node you check its neighbor's if they are visited then also you are checking them.Because of it your bfs code complexity can be roughly given as O(N*max_neighbours) as for each node you check all its neighbor's.Hope you got it what I want to say.

johnwick2b67

abhiaps

please someone tell me whats the problem in my code for C part div2 https://mirror.codeforces.com/contest/1362/submission/82531892

I am taking last significant bit and add ans corresponding to that bit and then subsequently remove last significant bit. Then repeat the process till n not reaches 0.

_-pratik_shukla-_

https://mirror.codeforces.com/contest/1361/submission/83105695 can someone help me with test case 6 of problem D Div 2

For Div 1E, does anyone know how you're supposed to keep track of the lone ancestor that has an edge from your subtree when there is only one such edge? I managed to get this working by keeping track of all ancestors that have an edge from $$$v$$$'s subtree and then using some gross small-to-large merging thing, but it seems that there is a way to do it without any such log^2 factor; I've been trying to read people's code but I'm not totally sure what the idea is that people are using.

dougalves

Solution to E is short and brilliant, i was struggling in anything like this but i haven't find the solution, thank you by the editorial.

PremKapshi1234

4 years ago, # |

Don't you think for problem D(div 2) test case 40:-

4 4

1 3

2 4

1 2 2 3

ans should be 1 2 3 4

but judge shows ans to be -1......

why?

goingOn

emorgan

3 years ago, # |

Div2 B can be solved in O(nlogn) time with a segment tree and hashing. Surprised nobody noticed this yet! Code: 126126788

Brief explanation: let S be the upper bound on elements, then consider the binary string s of length S, where s[i] == '1' means the set contains i. Build a segment tree on that string, where each node stores a polynomial hash of the corresponding substring of s. Then the root node of that segment tree stores the hash of the whole string.

To permute the string according to the xor thing in the problem with input k, we can rebuild the segment tree where some nodes treat their children as swapped when computing the hash (the less significant bits of k correspond to nodes closer to the root). Now here is the important part: if the segtree is currently permuted according to some k, then to switch to having it be permuted according to k+1, we don't need to touch any of the nodes corresponding to a bit on the prefix where k and k+1 match -- for example if we are transitioning from 00101 to 00110, then we can ignore the last three layers of nodes and only process the two layers closest to the root.

Now say we run through all possible k from 1 to S-1 -- each time we will check if the current root hash matches the root hash for the original, unpermuted string. If it does then we found a candidate answer. To analyze the complexity, let's think about how many times each node gets recalculated. Each leaf get recalculated 1 time, the nodes right above the leaves get recalculated 2 time, the next layer 4 times, then 8, ..., and the root gets recalculated S times. Each layer contributes S to the answer, and there are log(S) layers, so the total runtime is O(SlogS).

seiko.iwasawa

Good idea! We can also build another segment tree.

Let's recursively build an array of hashes of all possible sets that we can get. So, $$$tree[v][x]$$$ stores the hash of $$$s \oplus x$$$, where $$$0 \le x < 2^h$$$, where $$$h$$$ is the height of the vertex $$$v$$$. Note that we can do it without recursion like in merge sort.

Code: 126167404.

henriqueramos.qs

8 months ago, # |

An alternative solution to "1361B — Johnny and Grandmaster" would be sticking with the trivial solution

vector<int> v(n);
sort(v.begin(),v.end(),greater<int>());

sum =0;

for(int c:v){
    if(sum==0){
        sum+=fastexp(p,v[i]);
    }else{
        sum-=fastexp(p,v[i]);
    }   
    sum%=MOD;
}

But using two mods to ensure that a sequence of operations using the potencies of p will not result in a multiple of MOD and thus be misinterpreted as 0.

AC code: https://mirror.codeforces.com/contest/1361/submission/258625992

Okrut's blog

This was my solution(82558750) but last test case (test case 8) input 2**59 — 1 it failed

include

include<bits/stdc++.h>