1 solutions
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0
#include<bits/stdc++.h> using namespace std; bool k[10005]; bool d[21000]; int a,b,ans; int main(){ cin>>a; while(a--){ cin>>b; k[b]=1; } for(int i=1;i<=10005;i++){ for(int j=1;j<=10005;j++){ if(i!=j&&k[i]&&k[j]&&k[i+j]&&d[i+j]==0){ ans++; d[i+j]=1; } } } printf("%d",ans); return 0; }
- 1
Information
- ID
- 115
- Time
- 1000ms
- Memory
- 128MiB
- Difficulty
- 2
- Tags
- # Submissions
- 55
- Accepted
- 8
- Uploaded By