Noobs Round #2 (Div. 4) Editorial

№	Пользователь	Рейтинг
1	tourist	3985
2	jiangly	3741
3	jqdai0815	3682
4	Benq	3529
5	orzdevinwang	3526
6	ksun48	3489
7	Radewoosh	3483
8	Kevin114514	3442
9	ecnerwala	3392
9	Um_nik	3392

№	Пользователь	Вклад
1	cry	169
2	maomao90	162
2	Um_nik	162
2	atcoder_official	162
5	djm03178	158
6	-is-this-fft-	157
7	adamant	155
8	awoo	154
8	Dominater069	154
10	nor	150

Contest link : https://mirror.codeforces.com/gym/102942

Announcement link : https://mirror.codeforces.com/blog/entry/87209

Finally, The round is over smoothly :) We hope everyone had fun and enjoyed the contest!

Video Editorial by SecondThread.
Screencast by AnandOza (start 10 minutes after the end of the contest).
Video Editorial by Bojack_human. He is a new youtuber. So, please support him (:

A. Directional Move

Tutorial

Solution

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

#define ll long long int
#define pb push_back
#define mp make_pair
#define all(x) x.begin(),x.end()

int main()
{
    int test,n;
    cin>>test;
    while(test--){
	string side="ESWN",p;
	cin>>n>>p;

	ll idx=0;
	for(ll i=0;i<n;i++){
		if(p[i]=='0') idx++;
		else idx--;
		if(idx==-1) idx=3;
		else if(idx==4) idx=0;
	}
	 cout<<side[idx]<<endl;
    }   
    return 0;
}

Tester Solution

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

#define      PSB                 push_back
#define      ll                 long long
#define      FastIO      ios::sync_with_stdio(0);cin.tie(0); cout.tie(0);
constexpr ll mod = 1e9  + 7;
const ll N=3e5+5;

int main(){
  FastIO
  
  int test;
  cin>>test;
  while(test--){
      ll cnt=0,n;
      string s;
      cin>>n>>s;
      for(ll i=0;i<n;i++){
        if(s[i]=='0') cnt++;
        else cnt--;
      }
      cnt%=4;
      if(cnt<0) cnt+=4;

      if(!cnt) cout<<"E"<<endl;
      else if(cnt==1) cout<<"S"<<endl;
      else if(cnt==2) cout<<"W"<<endl;
      else cout<<"N"<<endl;
  }

  return 0;
}

B. Make All Odd

Tutorial

Solution

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

#define      PSB         push_back
#define      ll          long long
#define      FastIO      ios::sync_with_stdio(0);cin.tie(0); cout.tie(0);
constexpr ll mod = 1e9  + 7;
const ll N=3e6+5;

int main(){
  FastIO
  
  int test;
  cin>>test;
  while(test--){
      int n;
      cin>>n;
      vector<int> a(n);
      int cnt=0; bool ck=false;
      for(int i=0;i<n;i++){
        cin>>a[i];
        if(a[i]%2) ck=true;
        else cnt++;
      }

      if(!ck) cout<<"-1"<<endl;
      else cout<<cnt<<endl;
  }

  return 0;
}

C. Team

Tutorial

Solution

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

#define      PSB         push_back
#define      ll          long long
#define      FastIO      ios::sync_with_stdio(0);cin.tie(0); cout.tie(0);
constexpr ll mod = 1e9  + 7;
const ll N=3e6+5;

int main(){
  FastIO
  
  int test;
  cin>>test;
  while(test--){
      int n,k,team=0;
      cin>>n>>k;
      vector<int> a(n);
      for(int i=0;i<n;i++) cin>>a[i];
      sort(a.begin(),a.end());

      for(ll l=0,r=n-1;l<=r; ){
         if(a[r]>=k){
            ++team; --r;
         }
         else if(a[r]+a[l]>=k && l!=r){
            ++team; ++l; --r;
         }
         else l++;
      }
      cout<<team<<endl;
  }

  return 0;
}

Tester Solution

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

#define      PSB         push_back
#define      ll          long long
#define      FastIO      ios::sync_with_stdio(0);cin.tie(0); cout.tie(0);
constexpr ll mod = 1e9  + 7;
const ll N=3e6+5;


int main(){
  FastIO
  
  int test;
  cin>>test;
  while(test--){
      int n,m,a;
      cin>>n>>m;
      multiset<ll> st;
      for(ll i=0;i<n;i++){
        cin>>a;
        st.insert(a);
      }
      ll ans=0;

      for(ll i=n-1;i>=0;i--){
        ll bal=*st.rbegin();
        st.erase(--st.end());
        if(bal>=m){
            ans++; continue;
        }
        ll baki=m-bal;
        auto xx=st.lower_bound(baki);
        if(xx==st.end()) break;
        st.erase(xx);
        ans++;
        if(st.size()==0) break;
      }
      cout<<ans<<endl;
  }

  return 0;
}

D. XOR Game

Tutorial

Solution

#include <bits/stdc++.h>
using namespace std;
 
#define      PSB                 push_back
#define      ll                 long long
#define      FastIO      ios::sync_with_stdio(0);cin.tie(0); cout.tie(0);
constexpr ll mod = 1e9  + 7;
const ll N=3e5+5;
 
int main(){
  FastIO
  
  int test;
  cin>>test;
  while(test--){
      int a,b;
      cin>>a>>b;
      if((a^b)==(a+b)) cout<<"No"<<endl;
      else cout<<"Yes"<<endl;
  }
  return 0;
}

Tester Solution

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

#define ll long long int
#define pb push_back
#define mp make_pair
#define all(x) x.begin(),x.end()

int main()
{	
    int test;
    cin>>test;
    read:
    while(test--){
        ll a,b;
        cin>>a>>b;

        while(a && b){
            if(a%2 && b%2){
                cout<<"Yes"<<endl;
                goto read;
            }
            a/=2;
            b/=2;
        }
        cout<<"No"<<endl;
    }
	return 0;
}

E. Password

Tutorial

Solution

#include<bits/stdc++.h>

using namespace std;

const int N=1e5+5;
const int mod=1e9+7;

char s[N];
long long dp[N][10];

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);cout.tie(NULL);
    
    int test;
    cin>>test;
    while(test--){
        int i,j,l,n;
        cin >>n>> s+1;
        l=strlen(s+1);
        for(int i=0;i<=l;i++)
            for(int j=0;j<=10;j++)
              dp[i][j]=0;
        
        for(i=1;i<10;i++) dp[0][i]=1;
        for(i=1;i<=l;i++)
        {
            if(s[i] >= '1'  &&  s[i] <= '9')dp[i][s[i]-'0']=dp[i-1][s[i]-'0'];
            else
            {
                for(j=1;j<10;j++)dp[i][j]=dp[i-1][j];
            }
            for(j=2;j<10;j++)
            {
                dp[i][j]+=dp[i][j-1];
                dp[i][j]%=mod;
            }
        }
        cout << dp[l][9]<<endl;
    }
    return 0;
}

Tester Solution

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

#define ll long long int
#define pb push_back
#define mp make_pair
#define all(x) x.begin(),x.end()

int main()
{

    int test,n;
    cin>>test;
    read:
    while(test--){
        string s;
        cin>>n>>s;
        s='1'+s+'9';
        ll mod=1000000007LL,ans=1LL;

        for(ll i=0,j;i<(ll)s.size()-1;){
            for(j=i+1;j<(ll)s.size();j++)
                if(s[j]!='-') break;
            if(s[j]<s[i]){
                cout<<0<<endl;
                goto read;
            }
            ll diff=s[j]-s[i];
            ll d[10];
            for(ll k=0;k<10;k++) d[k]=1LL;

            for(ll k=i+1;k<j;k++){
                for(ll l=diff-1;l>=0;l--){
                    d[l]+=d[l+1];
                    d[l]%=mod;
                }
            }
            ans*=d[0];
            ans%=mod;
            i=j;
        }
        cout<<ans<<endl;
    }

	return 0;
}

Tester Solution 2

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int mod=1e9+7;
const int mx=1e6+5;
int Pow(int a,int b){
	int res=1;
	while(b){
		if(b&1)res=(res*1LL*a)%mod;
		a=(a*1LL*a)%mod;
		b>>=1;
	}return res;
}
int f[mx];

inline int mul(int a,int b){return (a*1LL*b)%mod;}
inline int ncr(int n,int k){
	return mul(f[n],Pow(mul(f[k],f[n-k]),mod-2));
}

int main(){
	f[0]=1;
	for(int i=1;i<mx;i++){
		f[i]=(f[i-1]*1LL*i)%mod;
	}
	
    int test;
    cin>>test;
    read:
    while(test--){
        int ans=1,n;
        string s; cin>>n>>s; s="1"+s+"9";  n=s.size();
        for(int i=0,l=-1;i<n;i++){
            if(s[i]>='0' && s[i]<='9'){
                if(l==-1){
                    l=i;
                    continue;
                }
                int a=s[l]-'0',b=s[i]-'0',k=b-a;
                if(k<0){
                    cout<<0<<endl;
                    goto read;
                }
                int len=i-l-1;
                if(len==0){
                    l=i;
                    continue;
                }
                // cout<<len<<' '<<k<<' '<<i<<endl;
                int aa=0;
                for(int K=0;K<=k;K++){
                    aa+=ncr(len+K-1,K);
                    if(aa>=mod)aa-=mod;
                }
                ans=mul(ans,aa);
                l=i;
            }
        }cout<<ans<<endl;
    }
	return 0;
}

F. Offer

Tutorial

Solution

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

#define      PSB         push_back
#define      ll          long long
#define      FastIO      ios::sync_with_stdio(0);cin.tie(0); cout.tie(0);
constexpr ll mod = 1e9  + 7;
const ll N=1e6+5;

int n, k;
int a[N], curf[N];

int main(){
    FastIO

    int test;
    cin>>test;
    while(test--){
        cin>>n>>k;
        vector<int> a(n);
        for(int i=0;i<n;i++) cin>>a[i],curf[a[i]]=0;

        int cost = 0, ans = 0;
        int l = 0;
        for(int r = 0; r < n; r++)
        {
            if(curf[a[r]] == 0)
                cost += a[r];
            curf[a[r]]++;
            while(cost > k)
            {
                curf[a[l]]--;
                if(curf[a[l]] == 0)
                    cost -= a[l];
                l++;
            }
            ans = max(ans, r - l + 1);
        }
        cout<<ans<<endl;
    }

  return 0;
}

Tester Solution

//Author: AnandRaj doubleux
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vpii;
typedef pair<ll,ll> pll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<pll> vpll; 

#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define test() int t;cin>>t;while(t--)
#define all(v) v.begin(),v.end()
#define prinp(p) cout<<p.first<<" "<<p.second<<endl
#define prinv(V) for(auto v:V) cout<<v<<" ";cout<<endl
#define take(V,f,n) for(int in=f;in<f+n;in++) cin>>V[in]
#define what(x) cerr<<#x<<" = "<<x<<endl
#define KStest() int t,t1;cin>>t;t1=t;while(t--)
#define KScout cout<<"Case #"<<t1-t<<": "
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int MOD = 1e9+7,MAX = 1e6+5;
const ll INF = 1e18+5;

int main()
{
    int test;
    cin>>test;
    while(test--){
        int n,k;
        cin>>n>>k;
        vi A(n);
        take(A,0,n);
        int ans = 0;
        int low = 0,high = n;
        while(low<=high)
        {
            int mid = low + (high-low)/2;
            bool can = false;
            ll sum = 0;
            vi M(MAX);
            for(int i = 0;i<mid;i++)
            {
                M[A[i]]++;
                if(M[A[i]]==1) sum+=A[i];
            }
            if(sum<=k)
            {
                ans = mid;
                low = mid+1;
            }
            else
            {
                for(int i=mid;i<n;i++)
                {
                    M[A[i-mid]]--;
                    if(M[A[i-mid]]==0) sum -= A[i-mid];
                    M[A[i]]++;
                    if(M[A[i]]==1) sum += A[i];

                    if(sum<=k) can=true;
                }
                if(can)
                {
                    ans = mid;
                    low = mid+1;
                }
                else high = mid-1;
            }
        }
        cout<<ans<<endl;
    }
}

-If have any more query, then please comment.

#include <bits/stdc++.h> using namespace std; template <typename A, typename B> string to_string(pair<A, B> p); string to_string(const char& ch) { return "'" + string(1, ch) + "'"; } string to_string(const string& s) { return '"' + s + '"'; } string to_string(const char* s) { return to_string((string) s); } string to_string(bool b) { return (b ? "true" : "false"); } string to_string(vector<bool> v) { bool first = true; string res = "{"; for (int i = 0; i < static_cast<int>(v.size()); i++) { if (!first) { res += ", "; } first = false; res += to_string(v[i]); } res += "}"; return res; } template <typename A> string to_string(A v) { bool first = true; string res = "{"; for (const auto &x : v) { if (!first) { res += ", "; } first = false; res += to_string(x); } res += "}"; return res; } template <typename A, typename B> string to_string(pair<A, B> p) { return "(" + to_string(p.first) + ", " + to_string(p.second) + ")"; } #define output cout void debug_out() { output << endl; } template <typename Head, typename... Tail> void debug_out(Head H, Tail... T) { output << " " << to_string(H); debug_out(T...); } #ifndef ONLINE_JUDGE #define dbg(...) output << "[" << #__VA_ARGS__ << "] :", debug_out(__VA_ARGS__) #else #define dbg(...) 42 #endif #define ll int64_t #define ull uint64_t #define lld long double #define FIO ios_base::sync_with_stdio(false);cin.tie(NULL); #define pb push_back #define eb emplace_back #define ff first #define ss second #define vt vector #define vll vt<ll> #define pll pair<ll,ll> #define vpll vt<pll> #define vvll vt<vll> #define all(v) v.begin(),v.end() #define rall(v) v.rbegin(), v.rend() #define FOR(i,n) for(ll i=0;i<n;i++) #define ffo(i,a,b) for(ll i=a;i<=b;i++) #define rfo(i,a,b) for(ll i=a;i>=b;i--) #define space cout<<"\n\n"; #define endl '\n' // comment this line in interactive prob template <typename T> using mxpq = priority_queue<T>; template <typename T> using mnpq = priority_queue<T, vt<T>, greater<T>>; #define fps(x,y) fixed<<setprecision(y)<<x #define merg(a,b,c) set_union(a.begin(),a.end(),b.begin(),b.end(),inserter(c,c.begin())) #define mmss(arr,v) memset(arr,v,sizeof(arr)) template<typename T1, typename T2> istream& operator>>(istream& in, pair<T1, T2> &a) {in >> a.ff >> a.ss; return in;} template<typename T1, typename T2> ostream& operator<<(ostream& out, pair<T1, T2> a) {out << a.ff << " " << a.ss; return out;} template<typename T, typename T1> T amax(T &a, T1 b) {if (b > a) a = b; return a;} template<typename T, typename T1> T amin(T &a, T1 b) {if (b < a) a = b; return a;} int dx[4] = { -1, 0, 1, 0}; int dy[4] = {0, 1, 0, -1}; // up, right, down, left const double PI = 3.14159265359; const ll INF = 0x3f3f3f3f3f3f3f3fll; const ll N = 1e6 + 6; const ll MAX_SIZE = 2e6 + 6; const ll mod = 1e9 + 7; ll powerM(ll x, ll y, ll M = mod) { // default argument ll v = 1; x = x % M; while (y > 0) {if (y & 1)v = (v * x) % M; y = y >> 1; x = (x * x) % M;} return v; } ll power(ll x, ll y) { ll v = 1; while (y > 0) {if (y & 1)v = v * x; y = y >> 1; x = x * x;} return v; } int largest_bit(long long x) { // based on 0-indexing return x == 0 ? -1 : 63 - __builtin_clzll(x); } const ll maxN = 1e6 + 9; void preSolve() { } void solve() { ll n, k; cin >> n >> k; vll a(n); FOR(i, n) { cin >> a[i]; } ll i = -1, j = 0, cost = 0; vll hash(maxN, 0); ll ans = 0; while (i < n) { if (i + 1 < n) { ++i; if (hash[a[i]] == 0) { cost += a[i]; hash[a[i]]++; // } else { hash[a[i]]++; } } // while (cost > k && j <= i) { if (hash[a[j]] > 1) { hash[a[j]]--; ++j; } else { hash[a[j]]--; cost -= a[j]; ++j; } } ans = max(ans, i - j + 1); if (i + 1 == n) break; } cout << ans << endl; } int main() { #ifdef LOCAL freopen("in1.txt", "r", stdin); freopen("out1.txt", "w", stdout); #endif FIO; preSolve(); int testcases = 1; cin >> testcases; for (int caseno = 1; caseno <= testcases; ++caseno) { // cout << "Case #" << caseno << ": "; solve(); } return 0; }

Комментарии (67)

Показать архивные | Написать комментарий?

humlefool

4 года назад, # |

thanks for this contest and hope for more contest like this.

→ Ответить

Hz_

4 года назад, # ^ |

Rudro25 what can be the difficulty of problems E and F?

RD_TheCoder

My guess: E(1400-1500) and F(1500-1600)

dush1729

← Rev. 2 →

Great round. For D, if x and y have no common bits then answer is NO else YES. So this will also work

if(x & y) cout << "Yes";
else cout << "No";

MohamedAboOkail

this will also work!

cout << ((a | b) == (a + b) ? "No" : "Yes") << "\n";

Sakhiya07

Both are same!

a + b = (a | b) + (a & b)

Letscode_sundar

thank you for this . I came up with the same logic , but didn't know how to proceed with bit manipulations.

So used the basic method to calculate every bit by n%2 and check whether both of them are 1 or not

if(a%2 && b&2)
{
cout<<"Yes"
return;
else
a=>>1;
b=>>1;
}

Loser_

Thanks for the round and quick editorial

welcome sir!

rahul_2

Thanks for the round waiting for the next one

welcome sir! i will try (:

Ankit_stcet

Hoping to solve upto D in div2 contest also some day. btw thanks for this contest and hope for more contest like this.

welcome sir :D i will try soon for the next one.

Aaste123

Welcome, Sir! Hope to see you next time also

SecondThread

+15

Nicely done, thanks for the contest!

Rudro25

+13

Thank you sir, for the support. orz

Saheel21

E was really a good problem for learning dp, thanks for the round.

hijibijee

We can also solve E with combinatorics. Combination with repetition to be exact. Suppose the string is like, 1----5. We can place any digit from this, {1,2,3,4,5} set in increasing order in 4 place between 1 and 5.

Let, n = number of place to fill. m = size of the set. Possible ways will be (n+m-1) Choose n.

For practice: https://mirror.codeforces.com/contest/1288/problem/C

mercurist

I thought on similar lines but wasn't able to get my code working as intended. I'd appreciate it if anyone out here could look into the following submission.

Link: https://ideone.com/U8cjmP

Try checking the validity of the result explicitly.

Matuagkeetarp

The problem is with the size of your fac array which is 100005 but according to the problem n<=10^6 which makes n+k-1<=100009(and it is less than 100005).

I changed 100005 to 100020 and it got Accepted

Silly me! Thank you so much!

Thanks for the formula, I got stuck at this and couldn't find this formula. Can you please provide me an explanation or a link where this formula can be find? Thanks for the given problem too.

alwyn

Stars and bars

Imagine you need to fill a sequence of integers of size $$$10$$$ using only $$$1$$$, $$$2$$$, or $$$3$$$ so that the sequence is non-decreasing. As they are non-decreasing, all $$$1$$$'s have to appear in front of all $$$2$$$'s and so on, so we don't need to care about the order of these integers and we need only the number of occurences of each $$$1$$$, $$$2$$$, and $$$3$$$. Thus, we need to find number of solution of: $$$count_1 + count_2 + count_3 = 10$$$, with $$$count_0, count_1, count_2 \geq 0$$$.

Check this out: https://www.youtube.com/watch?v=4_je4mXUCGA Hope it will be helpful. I learned it from Coursera. Someone uploaded the video in YouTube too.

dgupta0812

Consider this example:

1 — — — 6

In this we have to fill 3 places and to fill those 3 places we have 6 elements. Now observe that we only need to choose the elements with which we want to fill the gaps. We don't need to decide the order as that is fixed.

Qg3

I also tried to solve it with combinatorics. 105596767 This is the code it's failing test case 4. So can you help? if you find something. Thanks in advance.

Your solution will fail for (n+m-1) > 10e5 cases. Calculate factorials up to 10e6.
check this comment: https://mirror.codeforces.com/blog/entry/87253?#comment-754156

Ok Done :)

can you please explain the formula.how m+n-1

Nich728

Thanks for the contest, had fun doing it!

Welcome sir!

Tarun_19

+14

Please make the submissions public, or atleast the testcases. It helps a lot in debugging silly mistakes.

I already told this secondthread sir. May be he will open soon , if possible sir.

pj423

Could anyone explain F in a more detailed way? It would be much appreciated.

It's just 2 pointer approach, we can keep on increasing our right pointer while maintaining frequency and cost, until the cost becomes greater than k. At this point we can start incrementing our left pointer and decreasing the cost. The number of elements will right-left+1. Now we keep on repeating the prcoess.

Code

If in problem D, we were asked to maximize the score of Bob, how will we solve it ?

blank_manash

I think You can iterate from the highest to lowest bit, and try to set it, if possible.

AishwaryaR

The contest was great! Although I had come up with a similar idea as editorial, I had a hard time proving D during contest. I still can't wrap around why case 2 always is true. Could anyone please explain?

Hope This comment helps you.

sk_arman

Nice Contest..problem set was really good.

Thank you bhai :D

ksandr1v

Very nice contest! Want more! Especially thanks for the good E task.

Thank you sir!

pulsatio

The F can also be solved using the bit + binary lifting technique.

fluffycake

But isn't it overkill?

can you please add more test cases? bacause it is just checking with one test case and getting accepted?

so now i don't even know whether my solution is correct or not?

You have the permission to see only 1 test case, sir. May be it's the 'gym' contest rule.

dedsec_29

For C, why is the following approach incorrect? Answer is at least equal to elements >=k. Now put other elements in a list and sort it. Now iterate over every element, for every a[i], use lower_bound() on k-a[i] in range [i, list.size()] to find if such element exists. If it does, mark it used and increment answer. Why does this give a WA?

#include <bits/stdc++.h>
using namespace std;
//#pragma GCC optimize "trapv"
#define ll long long

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t; cin>>t;
    while (t--) {
        int n,k; cin>>n>>k;
        int ans = 0;
        vector<int> v;
        v.reserve(n/2);
        for (int i=0;i<n;i++) {
            int r; cin>>r;
            if (r>=k) ans++;
            else v.push_back(r);
        }
        sort(v.begin(), v.end());
        int len = v.size();
        vector<bool> used(len);
        for (int i=0;i<len-1;i++) {
            if (used[i]) continue;
            int w = k - v[i];
            int pos = lower_bound(v.begin()+i+1, v.end(), w) - v.begin();
            if (pos==len) continue;
            used[pos] = true;
            ans++;
        }
        cout<<ans<<"\n";
    }
    return 0;
}

can you please provide the code by spoiler! no permission to see other submission :(

Sure! My bad

Failed test case: 1 10 5 1 3 2 1 4 3 4 2 3 1

Expected: 4 but found 6.

May be the problem is — When you find a pos by lower_bound it can be also used before but you not checked that. may be you can use set and remove every used element then you can use lower_bound easily.

Ah right, I think the problem exactly is what you pointed out, I should definitely be removing those used values from my list. Thanks! Orz

LITT_UP

Let's consider a variation of problem F. Now in the subsegment [l,r] instead of making the cost of all repeating elements in that subsegment zero if we were allowed to take only one of the repeating elements in that subsegment and make its cost equal to zero. Rest all the conditions remain same as mentioned in the problem F. Then can someone please provide an insight on how we could have solved it.

procodgok

Hello , Thanks for this Contest . Can you Please Update the Settings so that we can we view all the test case , My Code for F gave wrong answer on test case 3 but I cannot access that test case . Thanks .

secondthread sir told that's may be impossible sir. but you can provide your code, i can check that by using polygon which on failed (:

Vik

We need more contests like this. Contests like today's destroy us and you pick us up. Thanks for organizing this man!

Eckchart

← Rev. 3 →

So, my understanding of the solution of $$$D$$$ is as follows:

If $$$a$$$ and $$$b$$$ are given such that there are no common bits between them, the answer is $$$NO$$$, because there are no 2 integers $$$1 \leq c \leq a,$$$ $$$1 \leq d \leq b$$$ s.t. their xor will be strictly greater than the xor between $$$a$$$ and $$$b$$$.

If $$$a$$$ and $$$b$$$ are given such that there are common bits between them, the answer is $$$YES$$$, because we can always find 2 integers $$$1 \leq c \leq a,$$$ $$$1 \leq d \leq b$$$ s.t. their xor will be strictly greater than the xor between $$$a$$$ and $$$b$$$.

But I don't understand why this should always be true (i.e. for 2 integers $$$a$$$ and $$$b$$$ s.t. there are common bits between them, why is it that we can always find 2 integers $$$1 \leq c \leq a,$$$ $$$1 \leq d \leq b$$$ which don't have common bits between them, s.t. their xor will always be strictly greater than the xor of $$$a$$$ and $$$b$$$), can anyone please provide a more detailed explanation of this or a proof?

Edit: deleted some erroneous stuff I thought was correct..; also, thanks a lot @doubleux, that last paragraph (especially) really made it very clear and obvious that the "strategy" for the solution of $$$D$$$ would always work.

doubleux

Notice, that the maximum Xor between any two numbers is always less than their sum, the maximum arising when no two bits are common.

Thus, if no bits are common, Xor of a and b is a+b, which is maximum that you can get using numbers less than them. Therefore no pair possible in this case.

If the do have common bits, a Xor b will be 0 at that position. Now choose c same as a, while for d, change the bit at common position as 0. This will make that bit 1 in their bitwise Xor, thus making it larger.

Hex1729

I used the same logic and approach in F but got TLE, why Python why :'-)

Problem F submission

Great tasks tho!

boy_of_kutbilim

I used the chorus property. a^b=c -> a^c=b so (c^d = (a^b)+somenum) -> (c^((a^b)+somenum)=d) And did 1<=somenum<=10. But why +10 was enough I don't understand. can open my eyes?

this is my submission.107902945

dhruv7888

← Rev. 5 →

IGNORE

drinkWater.

3 года назад, # |

Video tutorial for problem D: https://youtu.be/y2ERMJFsyjg

sshivendra764

please conduct more these type of contest . This is very helpful for beginners.

MishkatIT

10 месяцев назад, # |

Thanks for the contest!

Блог пользователя Rudro25