1001_A+B for Matrices

//題目1001：A+B for Matrices
//時間限制：1 秒內存限制：32 兆特殊判題：否提交：20447解決：8174
//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

#include "stdafx.h"
#include "stdio.h"
#include "string.h"

#define MAX 11

int a[MAX][MAX];
int column[MAX];
int row[MAX];

int main()
{
    //freopen("1001_data.in","r",stdin);
    //freopen("1001_data.out","w",stdout);
    int n,m,temp,count;
    while(scanf("%d %d",&m,&n) == 2 && m!=0)
    {
        for(int i = 0;i<MAX;i++)
        {
            column[i]=1;
            row[i]=1;
        }
        count = 0;
        for(int i = 0;i<m;i++)
            for(int j=0; j<n ;j++)
            {
                scanf("%d",&a[i][j]);
            }
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                scanf("%d",&temp);
                a[i][j] += temp;
                if(a[i][j]!=0)
                {
                    row[i] = 0;
                    column[j] = 0;
                }
            }
            count += row[i];
        }

        for(int j = 0;j<n;j++)
        {
            count += column[j];
        }
        printf("%d\n",count);
    }
    return 0;
}