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这个,,在杭电上也能A(考虑了全为负的情况)#include<iostream> using namespace std; int main() { int A[200][200]={0}; int N,i,j,k; while(~scanf("%d",&N)) { for(i=0;i<N;i++) { for(j=0;j<N;j++) { cin>>A[i][j]; } } int B[200][200]={0}; int max=A[0][0]; int s=0; for(j=0;j<N;j++) { for(i=0;i<N;i++) { B[i][j]=s+A[i][j]; s=B[i][j]; } s=0; } s=0; for(k=0;k<N;k++) { for(i=k;i<N;i++) { for(j=0;j<N;j++) { if(i==k) s+=A[i][j]; else { if(k==0) s+=B[i][j]; else { if(k==0) s+=B[i][j]; else s+=B[i][j]-B[k-1][j]; } } if(s>max && s!=0) max=s; if(s<0) s=0; } s=0; } } cout<<max<<endl; } return 0; } Followed by: Post your reply here: |
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