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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：

1 5

代码：

#include
     
      #include
      
        int main(){int a,b,x,y;int m[15][15],n[15][15],z[15][15];	int i,j;	while(scanf("%d%d",&a,&b)!=EOF)	{                                   if(a==0||b==0)                                   {                                   continue;}		for(i=1;i<=a;i++)	{		for(j=1;j<=b;j++)		{			scanf("%d",&x);			m[i][j]=x;		}			}	int k,v;	for(k=1;k<=a;k++)	{		for(v=1;v<=b;v++)		{	scanf("%d",&y);			n[k][v]=y;		}			}for(i=1;i<=a;i++)	{		for(j=1;j<=b;j++)		{			z[i][j]=m[i][j]+n[i][j];		}}int result=0,num=0;for(i=1;i<=a;i++){	for(j=1;j<=b;j++)	{		result+=z[i][j];	}	if(result==0)	{		num++;	}	if(result!=0)	{		result=0;	}}for(k=1;k<=b;k++){	for(v=1;v<=a;v++)	{		result+=z[v][k];	}	if(result==0)	{		num++;	}	if(result!=0)	{		result=0;	}}printf("%d\n",num);}}
      
     

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