1 solutions
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3
#include<bits/stdc++.h> using namespace std; const int N=110; char g[N][N]; bool r[N],c[N]; int main() { int n,m; cin>>n>>m; for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { cin>>g[i][j]; if(g[i][j]=='#') { r[i]=1; c[j]=1; } } } for(int i=1;i<=n;i++) { if(r[i]==1) //说明第i行至少有一个# { for(int j=1;j<=m;j++) { if(c[j]==1) //说明第j列至少有一个# { cout<<g[i][j]; } } cout<<endl;//每行输出完了以后输出换行 } } return 0; }
- 1
Information
- ID
- 1642
- Time
- 2000ms
- Memory
- 1024MiB
- Difficulty
- 3
- Tags
- (None)
- # Submissions
- 54
- Accepted
- 14
- Uploaded By