九度OJ 1001：A+B for Matrices

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：17682

解决：7079

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 21 11 1-1 -110 92 31 2 34 5 6-1 -2 -3-4 -5 -60

样例输出：

15

来源：

2011年浙江大学计算机及软件工程研究生机试真题

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****************************************************************/