九度OJ 1001:A+B for Matrices
- 题目描述:
- 
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. 
- 输入:
- 
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. 
- 输出:
- 
For each test case you should output in one line the total number of zero rows and columns of A+B. 
- 样例输入:
- 
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 
- 样例输出:
- 
1 5 
思路:
本题的意思是计算两个矩阵的和中, 全0行和全0列的个数。
代码:
#include <stdio.h>
 
#define N 10
 
int main(void)
{
    int m, n, i, j;
    int a[N][N], b[N][N];
 
    while (scanf("%d", &m) != EOF)
    {
        if (m == 0)
            break;
 
        scanf("%d", &n);
        for(i=0; i<m; i++)
        {
            for(j=0; j<n; j++)
            {
                scanf("%d", &a[i][j]);
            }
        }
        int count = 0;
        for(i=0; i<m; i++)
        {
            int zerorow = 1;
            for(j=0; j<n; j++)
            {
                scanf("%d", &b[i][j]);
                a[i][j] += b[i][j];
                if (a[i][j] != 0)
                    zerorow = 0;
            }
            count += zerorow;
        }
        for(j=0; j<n; j++)
        {
            int zerocol = 1;
            for(i=0; i<m; i++)
            {
                if (a[i][j] != 0)
                    zerocol = 0;
            }
            count += zerocol;
        }
 
        printf("%d\n", count);
    }
 
    return 0;
}
/**************************************************************
    Problem: 1001
    User: liangrx06
    Language: C
    Result: Accepted
    Time:0 ms
    Memory:912 kb
****************************************************************/ 
                    
                 
                
            
         
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