codeforces-infinite sequence

#include<iostream>
using namespace std;
int main()
{
 long long a,b,c;
 cin>>a>>b>>c;
 if(c==0)
 { 
  if(a==b)
   cout<<"YES";
  else
   cout<<"NO";
 }
 else
 {
  if((b-a)%c==0 and (b-a)/c>=0)
  {
   cout<<"YES";
  }
  else
  {
   cout<<"NO";
  }
 }
 return 0;
}

Comments

Codeforces-Dubstep solution

#include < bits / stdc ++. h > using namespace std ; int main () { string str ; getline ( cin , str ); int index = str . find ( "WUB" ); int count = 0 ; while ( index !=- 1 ) { str . replace ( index , 3 , " " ); //str.erase(str.begin()+index,str.begin()+index+3); index = str . find ( "WUB" ); } cout << str ; return 0 ; }

Counting the number of strongly connected sub graph

#include<bits/stdc++.h> using namespace std; class graph { int v; list<int> *adj; public: graph(int v) { this->v=v; adj=new list<int>[v]; } void add_edge(int v,int w) { adj[v].push_back(w); adj[w].push_back(v); } void dfs(int start,bool visited[]); void counted(int start); }; void graph:: dfs(int start,bool visited[]) { visited[start]=true; for(auto it=adj[start].begin();it!=adj[start].end();it++) { if(!visited[*it]) { dfs(*it,visited); } } } void graph:: counted(int start) { bool *visited=new bool[v]; for(int i=0;i<v;i++) visited[i]=false; int count=0; visited[start]=true; for(int k=0;k<v;k++) { if(!visited[k]) { dfs(k,visited); count++; } } cout<<"count is "<<count<<endl; } int main() { int v; cout<<"E

codeforces-Anton and Danik

#include < bits / stdc ++. h > using namespace std ; int main () { long n ; string str ; cin >> n ; cin >> str ; long count = 0 , index ; index = str . find ( "A" ); while ( index !=- 1 ) { count ++; str . erase ( str . begin ()+ index , str . begin ()+ index + 1 ); index = str . find ( "A" ); } if ( count > n - count ) cout << "Anton" ; else if ( count < n - count ) cout << "Danik" ; else cout << "Friendship" ; return 0 ; }

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