题目 1001 A+B for Matrices 九度Online Judge

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

/*
这道题关键是读懂题目的意思：
题目大意是要求输入两个整数M,N，然后分别输入两个M行N列的矩阵
将两个矩阵相加。最后输出的是相加后的矩阵中全部是0的行数和列
数。
*/

#include <stdio.h>
int main()
{
int M,N;
int i,j,count,a;
int m[10][10];
while (scanf("%d %d",&M,&N)!=EOF&&M)
{
count=M+N;                            //最后输出的计数值。这是一个关键点，如果从零开始计数，则会很//困难；所以就用减的方法。
for (i=0;i<M;i++)
{
for (j=0;j<N;j++)
{
scanf("%d",&m[i][j]);
}
}
for (i=0;i<M;i++)
{
for (j=0;j<N;j++)
{
scanf("%d",&a);
m[i][j]+=a;               //直接对矩阵相加
}
}
for (i=0;i<M;i++)
{
for (j=0;j<N;j++)
{
if (m[i][j]!=0)
{
count--;
break;
}
}
}
for (i=0;i<N;i++)
{
for (j=0;j<M;j++)
{
if (m[j][i]!=0)
{
count--;
break;
}
}
}
printf("%d\n",count);
}
return 1;
}

/**************************************************************
Problem: 1001
User: Carvin
Language: C++
Result: Accepted
Time:0 ms
Memory:1020 kb
****************************************************************/