Skip to main content

All Solution Code of CodeChef May Long Challenge 2021

competitve programming solution here...

solution code of Modular Equation problem number-5 of may long challenge codechef

 Modular Equation Problem Code: CodeChef


Solution Code

#include<bits/stdc++.h>
using namespace std;
#define int long long int
#define endl "\n"

int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int t;cin>>t;
    while(t--)
    {
        int n, m;cin>>n>>m;
        int count_of_pair = 0;
        vector<int> modular_equation(n+11);
        for(int a = 2;a<=n;a++)
        {
            int x = m%a;
            count_of_pair += modular_equation[x];
            for(int b = x;b<=n;b+=a)
            {
                modular_equation[b]++;
            }
        }
        cout<<count_of_pair<<endl;
    }
    return 0;
}

Comments

Popular posts from this blog

solution code of Tree House problem number-6 of may long challenge codechef

 Tree House Problem Solution Code: CodeChef Solution Code #pragma   GCC   optimize (" Ofast ", " unroll-loops ") #include   <bits/stdc++.h> using   namespace   std ; #define   int   long   long   int #define   double   long   double using   pii   =   pair < int ,  int >; template  < typename   T > using   Prior   =   std :: priority_queue < T ,  vector < T >,  greater < T >>; #define   X  first #define   Y  second #define   eb  emplace_back #define   ALL ( x )  begin (x),  end (x) #define   RALL ( x )  rbegin (x),  rend (x) #define   fastIO ()  ios_base ::sync_with_stdio, cin. tie ( 0 ) mt19937_64  rng( chrono :: steady_clock :: now (). time_since_epoch (). count ()); inline   int   getRand ( int   L ,  int   R ) { ...

solution code of Tic Tac Toe problem number-4 of may long challenge codechef

 Tic Tac Toe Problem Solution Code:  CodeChef Solution Code #include <bits/stdc++.h> using   namespace   std ; #define   ll   long   long   int int   main () {      ll  t;     cin >> t;      while (t -- )     {          ll  cx  =   0 , co  = 0 , c_ =   0 ;          char  a[ 3 ][ 3 ];          for ( ll  i = 0 ;i < 3 ;i ++ )         {              for ( ll  j = 0 ;j < 3 ;j ++ )             {                 cin >> a[i][j];  ...