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 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

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！！