ICM Technex 2017 and Codeforces Round #400 (Div. 1 + Div. 2, combined) Editorial

№	Пользователь	Рейтинг
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

№	Пользователь	Вклад
1	cry	169
2	maomao90	162
2	Um_nik	162
4	atcoder_official	160
5	djm03178	158
5	-is-this-fft-	158
7	adamant	155
8	Dominater069	154
9	awoo	152
10	luogu_official	150

Tutorial is loading...

Code

#include <algorithm>
#include <bitset>
#include <cassert>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>

#ifdef PRINTERS
#include "printers.hpp"
using namespace printers;
#define tr(a)		cerr<<#a<<": "<<a<<endl;
#else
#define tr(a)    
#endif

#define ll          long long
#define pb          push_back
#define mp          make_pair
#define pii         pair<int,int>
#define vi          vector<int>
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define sz(x)       (int)x.size()
#define hell        1000000007
#define endl        '\n'
#define rep(i,a,b)	for(int i=a;i<b;i++)
using namespace std;

void solve(){
	string a,b;
	cin>>a>>b;
	vector<string>taken;
	int n;
	cin>>n;
	rep(i,0,n){
		cout<<a<<" "<<b<<endl;
		string c,d;
		cin>>c>>d;
		if(c==a)a=d;
		else b=d;
	}
	cout<<a<<" "<<b<<endl;
}

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int t=1;
	while(t--){
		solve();
	}
	return 0;
}

Tutorial is loading...

Code

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

int sieve[100005];

int main()
{
	int i, n, j;
	cin>>n;
	for(i=2; i<=n+1; i++)
	{
		if(!sieve[i])
			for(j=2*i; j<=n+1; j+=i)
				sieve[j]=1;
	}
	
	if(n>2)
		cout<<"2\n";
	else
		cout<<"1\n";

	for(i=2; i<=n+1; i++)
	{
		if(!sieve[i])
			cout<<"1 ";
		else
			cout<<"2 ";
	}

	return 0;
}

Tutorial is loading...

Code

#include <algorithm>
#include <bitset>
#include <cassert>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>

#ifdef PRINTERS
#include "printers.hpp"
using namespace printers;
#define tr(a)        cerr<<#a<<": "<<a<<endl;
#else
#define tr(a)    
#endif
#define ll          long long
#define pb          push_back
#define mp          make_pair
#define pii         pair<int,int>
#define vi          vector<int>
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define sz(x)       (int)x.size()
#define hell        1000000007
#define endl        '\n'
#define rep(i,a,b)    for(int i=a;i<b;i++)
using namespace std;

void solve(){
    int N,k;
    scanf("%d%d",&N,&k);
    ll x[N+1];
    x[0]=0;
    rep(i,0,N)scanf("%lld",&x[i+1]);
    partial_sum(x,x+N+1,x);
    if(k==1){
        ll ans=0;
        map<ll,int>m;
        for(int i=N;i>=0;i--){
            if(m.find(x[i]+1)!=m.end())
                ans+=m[x[i]+1];
            m[x[i]]++;
        }
        printf("%lld\n",ans);
    }
    else if(k==-1){
        ll ans=0;
        map<ll,int>m;
        for(int i=N;i>=0;i--){
            if(m.find(x[i]+1)!=m.end())
                ans+=m[x[i]+1];
            if(m.find(x[i]-1)!=m.end())
                ans+=m[x[i]-1];
            
            m[x[i]]++;
        }
        printf("%lld\n",ans);
    }
    else{
        ll ans=0;
        map<ll,int>m;
        for(int i=N;i>=0;i--){
            ll cur=1;
            while(true){
                if(abs(cur)>=1e15)break;
                if(m.find(x[i]+cur)!=m.end())
                ans+=m[x[i]+cur];
                cur*=k;
            }
            m[x[i]]++;
        }
        printf("%lld\n",ans);
    }
}

signed main(){
    int t=1;
    while(t--){
        solve();
    }
    return 0;
}

Tutorial is loading...

Code

#include <bits/stdc++.h>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
#define bitcount    __builtin_popcountll
#define mod 1000000007
#define N 1000009
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ss(s) cin>>s;
#define si(x)  scanf("%d",&x);
#define sl(x)  cin>>x;
#define pb push_back
#define mp make_pair
#define all(v) v.begin(),v.end()
#define S second
#define F first
#define ll long long
int n,m;
vector<int> arr[N];
int arr1[N];
vector<pair<int,int > > adj[100008];                    //max-size
void graph(int n){for(int i=1;i<=n;i++){
					int x = arr[i][0];
					int y = arr[i][1];
					adj[x].push_back(mp(y,arr1[i] ));                           //arr1[i] stores th color of the edge
					adj[y].push_back(mp(x,arr1[i]) );}

}
int col[N];
bool vis[N];
int main(){
//fast                          //uncomment on submission
cin>>n>>m;
assert(n>=2 && n<=100000);
assert(m>=2 && m<=100000);
for(int i =1;i<=n;i++)cin >> arr1[i];
for(int i = 1;i<=m;i++){
    int sz;
    cin >> sz;
    for(int j =0;j<sz;j++){
        int y ;
        cin >> y;
        arr[y].push_back(i);
    }
}
graph(n);
mem(col,-1);
mem(vis,false);
bool p = true;
for(int i = 1;i<=m;i++){
	if(!vis[i] && adj[i].size()){
		col[i] = 0;
		vis[i] = true;
		queue <int > q;
		q.push(i);

		while(q.size()){
			int node = q.front();
			q.pop();
	
			for(auto pi:adj[node]){
				int co = col[node];
				if(pi.S == 0){                       //if the edge is 0 then give the opposite color.
					co = 1 - co;

		}
		if(vis[pi.F] && col[pi.F] != co){
			p = false;
		}
		if(!vis[pi.F]){
			col[pi.F] = co;
			vis[pi.F] = true;
			q.push(pi.F);

		}
	}
}
}
}

if(!p)cout<<"NO";
else cout<<"YES";

}

Tutorial is loading...

Code

#include <bits/stdc++.h>
using namespace std;
#define ll          long long
#define MOD         1000000007
#define ll          long long
#define pb          push_back
#define pii         pair<int,int>
#define vi          vector<int>
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define endl        '\n'
#define PI          3.14159265359d
#define sz(x)       (int)x.size()
ll phi(ll n)
{
    ll i,res=n;
    for(i=2;i*i<=n;i++)
        if(n%i==0)
    {
        while(n%i==0)
            n/=i;
        res-=res/i;
    }
    if(n>1)
        res-=res/n;
    return res;
}
int main()
{
    ll n,k,res;
    cin>>n>>k;
    res=n;
    k=(k+1)/2;
    while(k--)
    {
        res=phi(res);
        if(res==1)
            break;
    }
    cout<<res%MOD;
    return 0;
}

Tutorial is loading...

Code

#include<bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d",&x)
#define slld(x) scanf("%lld",&x)
#define ss(x) scanf("%s",x)
#define ll long long
#define mod 1000000007
#define bitcount    __builtin_popcountll
#define pb push_back
#define fi first
#define se second
#define mp make_pair
#define pi pair<int,int>
pi arr[100005];
int n,m;
bool cmp(pi x, pi y)
{
	int a=min(x.se-x.fi-1,n-2-(x.se-x.fi-1));
	int b=min(y.se-y.fi-1,n-2-(y.se-y.fi-1));
	return a<b;
}
vector<int>nodes[100005];
vector<int>v[100005],v2[100005];
int de[100005],siz[100005];
void dfs1(int node, int par)
{
	siz[node]=1;
	for(int i=0;i<v[node].size();i++)
	{
		if(de[v[node][i]]==0&&v[node][i]!=par)
		{
			dfs1(v[node][i],node);
			siz[node]+=siz[v[node][i]];
		}
	}
}
int dfs2(int node, int par,int s)
{
	int maxi=0,maxinode;
	for(int i=0;i<v[node].size();i++)
	{
		if(de[v[node][i]]==0&&v[node][i]!=par&&siz[v[node][i]]>s/2)
		{
			return dfs2(v[node][i],node,s);
		}
	}
	//printf("%d\n",node );
	return node;
}
int decompose(int node)
{
	dfs1(node,-1);
	int cen=dfs2(node,-1,siz[node]);
	// printf("%d\n",cen );
	de[cen]=1;
	for(int i=0;i<v[cen].size();i++)
	{
		if(de[v[cen][i]]==0)
		{
			int x=decompose(v[cen][i]);
			//printf("%d\n", x);
			v2[cen].pb(x);
			v2[x].pb(cen);
		}
	}

	return cen;
}
int level[100005];
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int i,j,k;
    sd(n);
    sd(m);
    for(i=1;i<=m;i++)
    {
    	sd(arr[i].fi);
    	sd(arr[i].se);
    	if(arr[i].fi>arr[i].se)
    	{
    		swap(arr[i].fi,arr[i].se);
    	}
    }
    sort(arr+1,arr+m+1,cmp);
    set<int>s;
    for(i=1;i<=n;i++)
    	s.insert(i);
    k=0;
    for(i=1;i<=m;i++)
    {
    	if(arr[i].se-arr[i].fi-1<=n-2-(arr[i].se-arr[i].fi-1))
    	{
    		nodes[k].pb(arr[i].fi);
    		set<int>::iterator it=s.upper_bound(arr[i].fi);
    		while((*it)!=arr[i].se)
    		{
    			nodes[k].pb(*it);
    			it++;
    		}
    		nodes[k].pb(arr[i].se);
    		for(j=1;j<nodes[k].size()-1;j++)
    			s.erase(nodes[k][j]);
    	}
    	else
    	{
    		while(*s.begin()!=arr[i].fi)
    		{
    			nodes[k].pb(*s.begin());
    			s.erase(s.begin());
    		}
    		nodes[k].pb(arr[i].fi);
    		nodes[k].pb(arr[i].se);
    		int l=nodes[k].size();
    		set<int>::iterator it=s.upper_bound(arr[i].se);
    		while(it!=s.end())
    		{
    			nodes[k].pb(*it);
    			it++;
    		}
    		for(;l<nodes[k].size();l++)
    			s.erase(nodes[k][l]);
    	}
    	reverse(nodes[k].begin(), nodes[k].end());
    	k++;
    }
    while(s.size()!=0)
    {
    	nodes[k].pb(*s.begin());
    	s.erase(s.begin());
    }
   	reverse(nodes[k].begin(), nodes[k].end());
	k++;
    sort(nodes,nodes+k);
    map<pi,int>m1;
    for(i=0;i<k;i++)
    {
    	for(j=0;j<nodes[i].size()-1;j++)
    	{
    		if(nodes[i][j]==nodes[i][j+1]+1)
    			continue;
    		else if(m1.find(mp(nodes[i][j],nodes[i][j+1]))!=m1.end())
    		{
    			v[m1[mp(nodes[i][j],nodes[i][j+1])]].pb(i+1);
    			v[i+1].pb(m1[mp(nodes[i][j],nodes[i][j+1])]);
    			m1.erase(mp(nodes[i][j],nodes[i][j+1]));
    		}
    		else
    		{
    			m1[mp(nodes[i][j],nodes[i][j+1])]=i+1;
    		}
    	}
    	j=nodes[i].size()-1;
    	if(nodes[i][0]!=n||nodes[i][j]!=1)
    	{
    		if(m1.find(mp(nodes[i][0],nodes[i][j]))!=m1.end())
    		{
    			v[m1[mp(nodes[i][0],nodes[i][j])]].pb(i+1);
    			v[i+1].pb(m1[mp(nodes[i][0],nodes[i][j])]);
    			m1.erase(mp(nodes[i][0],nodes[i][j]));
    		}
    		else
    		{
    			m1[mp(nodes[i][0],nodes[i][j])]=i+1;
    		}
    	}
    }
    for(i=1;i<=n;i++)
    	level[i]=1e6;
    int cen=decompose(1);
	//printf("%d\n",cen );
	queue<int>q;
	i=0;
	q.push(cen);
	while(!q.empty())
	{
		j=q.size();
		while(j--)
		{
			int z=q.front();
			q.pop();
			level[z]=i;
			for(int l=0;l<v2[z].size();l++)
			{
				if(level[v2[z][l]]==1e6)
				{
					q.push(v2[z][l]);
				}
			}
		}
		i++;
	}
	for(i=1;i<=k;i++){
		//printf("%d %d\n",v2[i].size(),level[i] );
		printf("%d",level[i]+1 );
		if(i!=k)
		printf(" ");
	}
	printf("\n");
    return 0;
}

Tutorial is loading...

Code

#include<bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%lld",&x)
#define slld(x) scanf("%lld",&x)
#define ss(x) scanf("%s",x)
#define ll long long
#define mod 65536
#define bitcount    __builtin_popcountll
#define pb push_back
#define fi first
#define se second
#define mp make_pair
#define pi pair<int,int>
ll dp[17][65538][17];
char a[20];
ll b[20];
ll mo[1100005];
ll f(ll m,ll p,ll pos)
{
	if(dp[m][p][pos]!=-1)
		return dp[m][p][pos];
	if(pos==0)
	{
		if(((1<<m)&p))
			dp[m][p][pos]=1;
		else
			dp[m][p][pos]=0;
		return dp[m][p][pos];
	}
	ll j=0;
	for(ll i=0;i<=15;i++)
	{
		j+=f(max(m,i),mo[p*16+i],pos-1);
	}
	return dp[m][p][pos]=j;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
	ll t,n,i,j,k,l,r,m,p,ans1;
	mo[0]=0;
	for(i=1;i<=1100000;i++)
	{
		mo[i]=mo[i-1]+1;
		if(mo[i]==mod)
			mo[i]=0;
	}
	memset(dp,-1,sizeof dp);
	f(0,0,16);
	sd(t);
	while(t--)
	{
		ss(a);
		n=strlen(a);
		l=0;
		for(i=0;i<n;i++)
		{
			l=l*16;
			if(a[i]>='0'&&a[i]<='9')
				l+=a[i]-'0';
			else
				l+=a[i]-'a'+10;
		}
		ans1=0;
		if(l>0)
		{
			l--;
			j=l;
			for(i=0;i<n;i++)
			{
				b[i]=j%16;
				j/=16;
			}
			m=p=0;
			for(j=n-1;j>=0;j--)
			{
				for(k=0;k<b[j];k++)
					ans1-=dp[max(m,k)][mo[p*16+k]][j];
				m=max(m,b[j]);
				p=mo[p*16+b[j]];
			}
			ans1-=dp[m][p][0];
		}
		ss(a);
		n=strlen(a);
		r=0;
		for(i=0;i<n;i++)
		{
			r=r*16;
			if(a[i]>='0'&&a[i]<='9')
				r+=a[i]-'0';
			else
				r+=a[i]-'a'+10;
		}
		j=r;
		for(i=0;i<n;i++)
		{
			b[i]=j%16;
			j/=16;
		}
		m=p=0;
		for(j=n-1;j>=0;j--)
		{
			for(k=0;k<b[j];k++)
				ans1+=dp[max(m,k)][mo[p*16+k]][j];
			m=max(m,b[j]);
			p=mo[p*16+b[j]];
		}
		ans1+=dp[m][p][0];
		printf("%lld\n",ans1);
	}
    return 0;
}

Комментарии (29)

Показать архивные | Написать комментарий?

dmvs000

8 лет назад, # |

Can someone explain me the C Molly's chemical value. What would be the time complexity of the solution?

→ Ответить

selfcompiler

8 лет назад, # ^ |

← Rev. 2 →

O(log(base(k)[(10^14)])*log(n)*n)

Explanation:- 1. distinct power of K => log(base(k)[(10^14)]) 2. total distinct sum= total maximum distict cumulative sum is n , as we are using map for the search total searching complexity for required sum is log(n). 3. and array size is n . - See solution : Solution Link

dcordb

+26

I want to share an alternative solution to Problem D which uses 2-SAT. Let p and q be the switches of a door. If the door is locked (equal to 0) you need to get it to 1 applying p or q, this the same as doing $\text{[math]}$ . Similarly if the door is unlocked (equal to 1) you should do $\text{[math]}$ (because you want that the door doesn't change).

Then you should transform each of this expressions so that you get a conjunctive normal form, you can do it like this:

$\text{[math]}$
$\text{[math]}$

And now for each door you have one of this expressions and you can do 2-SAT with this.

_Mr.T_

+11

And then you can see that when you added 1 one-way edge, you also added its reverse as well thus you can easily solve this problem using disjoint set.

LucianIlea

6 лет назад, # ^ |

-8

Beautiful solution!

jianzuan

3 года назад, # ^ |

wow

ironsoul

-11

Можно пожалуйста вопрос про D. Предположим мы ходим по графу и разукрашиваем его, допустим текущая вершина покрашена в цвет 1. Далее мы продолжаем его разукрашивать, и каким-то образом идем по ребру веса 1 в эту же вершину и цвет не совпадает, мы сразу выводим "NO". Но разве мы не можем пойти по какому-нибудь другому пути, где их цвета совпадут? Почему это работает ?

KtlTheBest

Если ты покрасишь 1 вершину в один цвет, то есть только один способ покрасить оставшиеся вершины. Ты ходишь по вершинам и разукрашиваешь их. Если ты смог покрасить весь граф по выше описанному алгоритму, то ответ есть, иначе если ты не модешь разукрастить граф, то ответа не существует. Подумай над этим.

Taan

can you check my account? 2 problems were accepted but my rating don't change!

Your submissions were skipped because someone has the same code as you. Did you use ideone.com or other online IDEs? If yes, many people had a chance to copy-paste your solution(cheaters)

oh no, I give my code to my friends :|

akshit.lite

For problem C, using unordered_map gives TLE on test case 101, whereas it gives AC using maps. Aren't unorderd_maps supposed to give better runtime or is this because of the rehash or something ?

Because of collisions unordered_map works with complexity O(n) in worst case while worst case for simple map is O(logN)

shjwudp

776C — Molly's Chemicals why unordered_map cause TLE?

I just replace unordered_map with map.

AC code : http://mirror.codeforces.com/contest/776/submission/24956397
TLE code : http://mirror.codeforces.com/contest/776/submission/24956380

can someone tell me why? please.

solved now.
AC code : http://mirror.codeforces.com/contest/776/submission/24956769

kwangg

They do this often: http://mirror.codeforces.com/blog/entry/4898

ds_95

Problem D
Can someone please help me debug this code
The approach was following:
1. Find controlling switches for every room and join them with edge weight 0 if room is open else with edge weight 1.

2. If room is open it means sum of toggling times of switches controlling it, should be even,therefore edge of weight 0 is added and vice-versa. (Edges therefore denotes parity of sum of toggling times to open this room.)

3. After the graph is constructed look for an odd-weighted cycle in the graph.

4. Existence of an odd-weighted cycle would mean there are two different paritys of sum of toggling times of same pair of switches , which is not acceptable ,Hence ans is "No"

PalashTaneja

Can someone please explain the dictionary part in problem C.

Dey

You don't have to watch at every prefix before current position, with dictionary you can easy find number of required sums with log(n). Otherwise u will get TL, cause of O(log(10^14)*n^2)

ryuzakil

I dont undertand Problem C at all.How are we distinguishing whether we are getting subarray sum or subset sum?How are we making sure we are taking subsegment.I just dont get it. Please someone help and clarify in Laymen terms.

egor.okhterov

← Rev. 5 →

I also don't understand the method described in editorial, but I can tell about my solution =)

Let's say we have an array A = [2, 2, 2]. The total number of different subarrays looks like that (I will write all of them as pairs of indices [start, end]):
[0, 0], [0, 1], [0, 2], Subarrays starting with 0
[1, 1], [1, 2], Subarrays starting with 1
[2, 2]. Subarrays starting with 2

In total there are 3 + 2 + 1 = 3 × (3 + 1) / 2 = 6 different subarrays. For an array of size n there are n × (n + 1) / 2 subarrays — too much to handle them one by one. That is why we split them cleverly into groups.
As you can see from the picture above there are 3 different groups of subarrays. I will call the group of segments that starts with index 0 as group₀ = {[0, 0], [0, 1], [0, 2]}. The group of segments starting with index 1 as group₁ and the last group with a single subarray as group₂.

Why should we split them like that? This will allow us to reuse calculations for the first group₀ in the other groups in the following order:
group₀ ⟶ group₁ ⟶ group₂

That is, we first calculate the sums for all of the subarrays in the group₀ directly without any clever manipulations like that:

int A[3] = { 2, 2, 2 };
int group_0_sums[3] = { 0, 0, 0 };

group_0_sums[0] = A[0];
group_0_sums[1] = A[1] + group_0_sums[0];
group_0_sums[2] = A[2] + group_0_sums[1];

Now I will place these sums in a map<>, because we need to count how many times a certain sum has appeared among subarrays in group₀:

map<int, int> cnt;
for (int i = 0; i <= 2; i++)
  cnt[group_0_sums[i]]++;

And now we are ready to reuse them in group₁. First thing that we know is that in the best case, even if all of the sums of subarrays in group₁ will match our requirements (powers of k) the number of subarrays in group₁ is smaller than in group₀ by 1 (number of subarrays in group₀ is 3 and number of subarrays in group₁ is 2). That is why we have to get rid of one of the subarrays in cnt[].

What subarray sum we need to get rid of? That is subarray [0, 0], because it doesn't intersect with any of the subarrays in group₁:

cnt[group_0_sums[0]]--; // Get rid of subarray [0, 0].

Now cnt contains sums of the 2 subarrays: [0, 1] and [0, 2]. But they start at 0, so we cannot reuse them as if they are subarrays of group₁...
But we can notice one thing if we put remaining subarrays of group₀ with subarrays of group₁ in front of each other:
group₀: [0, 1], [0, 1, 2]
group₁: [1], [1, 2]

You see? They differ only in one thing: subarrays of goup₀ include the element A[0], but subarrays of group₁ are without it! So we can reuse all of the sums in cnt by shifting the value that we want to look up by A[0]. For example, if we want to check how many subarrays from group₁ have a sum = 4 we will do the following:

int shift = A[0];
int needed_sum = 4;

int subarrays_with_sum = cnt[needed_sum + shift];

Anand25021998

hey pixar!!Can you please post your solution link here...

Sure!

My original solution (I was describing it in my comment): 24936305

Solution based on the idea described in the editorial: 25086867

kirakira

Problem C Molly's Chemicals can be solved using divide & conquer: 24957185

First we compute prefix sum in O(N)

Divide the array into half, until it has 1 element left. We try to recursively find the desired segment sums from left half and right half (The segment TOTALLY fall into those half)

For conquer part, we add up the return value from left half, right half, and those segments which cross the middle line, which can be calculated as follows:

Let A[i] & A[i+1] be the elements across the middle line. Use O(N) to find all segment sums which segment end at A[i], named it LEFT. Similarly, use O(N) to find those starts with A[i+1], named it RIGHT
Use O(N lg N) to sort RIGHT. Then for each LEFT, for each K power, binary search RIGHT to find if there is a number which can add up to the number. This step use O(N*K*lg N)

Together we got T(N) = T(N/2) + O(NK lg N)

I tend to see K as a constant, so I think T(N) ~ O(N lg^2 N)

lahm07

In question C, can anyone please explain the author's solution..

Laika

Can someone help me find my mistake for Div2/D. I am solving it using 2-SAT and getting WA on test 24 25295970

rotavirus

Why doesn't author give a link https://en.wikipedia.org/wiki/Euler%27s_totient_function ?

arthurconmy

5 лет назад, # |

Wow, problem D is a bad problem

ycx2010

23 месяца назад, # |

Блог пользователя DeshiBasara