题目1001：A+B for Matrices

 

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：8897

解决：3653

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

 code：

#include <iostream>#include <cstdio>using namespace std;int main(){int m,n,i,j,count;int array1[10][20],array2[10][20];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",&array1[i][j]);}}for (i=0;i<m;i++){for (j=0;j<n;j++){scanf ("%d",&array2[i][j]);}}for (i=0;i<m;i++){for (j=0;j<n;j++){array1[i][j]=array1[i][j]+array2[i][j];}}bool flag;count = 0;for (i=0;i<m;i++){flag = true;for (j=0;j<n;j++){if (array1[i][j]!=0){flag=false;break;}}if (flag)count++;}for (j=0;j<n;j++){flag=true;for (i=0;i<m;i++){if (array1[i][j]!=0){flag=false;break;}}if (flag)count++;}printf("%d\n",count);}return 0;}

注意flag的位置，就easy了！！