2 solutions
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3
#include<bits/stdc++.h> using namespace std; const int N=110; double a[N][N]; int main() { int n; cin>>n; int m=0; double s=0; for (int i=1;i<=n;i++) { for (int j=1;j<=n;j++) { cin>>a[i][j]; } } for (int i=1;i<=n;i++) { for (int j=1;j<=n;j++) { if (i+j<n+1) //小于黄色部分 { s+=a[i][j]; m++; } } } cout<<fixed<<setprecision(1)<<s/m; return 0; }
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-1
#include<bits/stdc++.h> using namespace std; int a[100][100]; int main(){ int n,c=0; cin>>n; double p=0; for(int i=0;i<=n;i++){ for(int j=0;j<=n;j++){ cin>>a[i][j]; } } for(int i=0;i<n-1;i++){ for(int j=0;j<n-1;j++){ if(i+j==n){ p+=a[i][j]; c++; } } } p/=c; cout<<fixed<<setprecision(1)<<p; return 0; }
- 1
Information
- ID
- 71
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 1
- Tags
- (None)
- # Submissions
- 85
- Accepted
- 34
- Uploaded By