1001 A+B for Matrices

題目描述：

    This time, you are supposed to find A+B where A and B are two matrices, and then count the number of zero rows and columns.

輸入：

    The input consists of several test cases, each starts with a pair of positive integers M and N (≤10) which are the number of rows and columns of the matrices, respectively. Then 2*M lines follow, each contains N integers
in [-100, 100], separated by a space. The first M lines correspond to the elements of A and the second M lines to that of B.

    The input is terminated by a zero M and that case must NOT be processed.

輸出：

    For each test case you should output in one line the total number of zero rows and columns of A+B.

範例輸入：
2 2
1 1
1 1
-1 -1
10 9
2 3
1 2 3
4 5 6
-1 -2 -3
-4 -5 -6
0
範例輸出：
1

5

統計下行列之和為0的即可

#include <iostream>using namespace std;int main(){int m,n;int a[11][11];int b[11][11];int c[11][11];int sum;int answer;while(cin>>m){sum=0;answer=0;if(m!=0){cin>>n;for(int i=0;i<m;i++)for(int j=0;j<n;j++){cin>>a[i][j];}for(int i=0;i<m;i++)for(int j=0;j<n;j++){cin>>b[i][j];}for(int i=0;i<m;i++)for(int j=0;j<n;j++){c[i][j]=a[i][j]+b[i][j];}           for(int i=0;i<m;i++)   {   sum=0;   for(int j=0;j<n;j++)   {   sum+=c[i][j];   }   if(sum==0)   {   answer++;   }   }   for(int i=0;i<n;i++)   {   sum=0;   for(int j=0;j<m;j++)   {   sum+=c[j][i];   }   if(sum==0)   {   answer++;   }   }   cout<<answer<<endl;}else{break;}}return 0;}