時間限制：1 秒

記憶體限制：32 兆

特殊判題：否

提交：11539

解決：4694

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.

樣例輸入：

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

樣例輸出：

1
5

/**************************************************************
    Problem: 1001
    Language: C++
    Result: Accepted
    Time:0 ms
    Memory:1080 kb
****************************************************************/
#include <stdio.h>
#include<string.h>
const int MAX=1000;
int ans[MAX][15],cnt;
int main()
{
    int n,m,tmp,i,j;
    //freopen("G:\\in.txt", "r", stdin);
    while(scanf("%d%d",&n,&m)!=EOF){
        if(n==0) break;
        for(i=0;i<n;i++){  //輸入n行作為矩陣A
            for(j=0;j<m;j++)
                scanf("%d",&ans[i][j]);
        }
        for(i=0;i<n;i++){  //輸入n行作為矩陣A+B
            for(j=0;j<m;j++){
                scanf("%d",&tmp);
                ans[i][j]+=tmp;
            }
        }
        cnt=0;
        for(i=0;i<n;i++){
            for(j=0;j<m;j++){
                if(ans[i][j]!=0)
                    break;
            }
            if(j==m)
                cnt++;
        }
        for(i=0;i<m;i++){
            for(j=0;j<n;j++){
                if(ans[j][i]!=0)
                    break;
            }
            if(j==n)
                cnt++;
        }
        printf("%d\n",cnt);
    }   
    return 0;
}