A+B for Matrices

    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.

    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.

    For each test case you should output in one line the total number of zero rows and columns of A+B.

2 21 11 1-1 -110 92 31 2 34 5 6-1 -2 -3-4 -5 -60样例输出：
1
5
首先输入矩阵的行列数，然后输入矩阵的信息，然后将两个矩阵相加，分别计算相加之后的得到的新矩阵的每一行，每一列都为0的个数。
#include <iostream>using namespace std;int main(){	int m,n;	int a[100][100],b[100][100];	int sum=0,count;	while(1)	{		cin>>m;		if(m==0)			break;		cin>>n;		count=0;		for(int i=0;i<m;++i)			for(int j=0;j<n;++j)				cin>>a[i][j];		for(i=0;i<m;++i)			for(int j=0;j<n;++j)				cin>>b[i][j];		for(i=0;i<m;++i)		{			for(int j=0;j<n;++j)				sum+=a[i][j]+b[i][j];//计算行的和			if(sum==0)				++count;			sum=0;		}		for(int j=0;j<n;++j)		{			for(int i=0;i<m;++i)				sum+=a[i][j]+b[i][j];//计算列的和			if(sum==0)				++count;			sum=0;		}		cout<<count<<endl;	}	return 0;}/*2 21 11 1-1 -110 92 31 2 34 5 6-1 -2 -3-4 -5 -60*/