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
代码：
#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;
}
一直狗的提醒：
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，我自己写的太麻烦了，特别长
感觉你参考的话还是参考这些有用的比较好 我准备以后有答案的就不写了，多参考几个好的给你