九度OJ 题目1001：A+B for Matrices

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

我的答案：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
int main(){
//freopen("input.txt","r",stdin);
int m,n,a,i,j,k,p,q,s;//p是0行数，q室0列数,s是行累加和
int A [10][10];
int l [10];//列累加和数组
memset(A,0,sizeof(int)*100);
memset(l,0,sizeof(int)*10);
while(cin>>m && m)
{
cin >> n;
p=0,q=0;//每次2组过后，p,q重置
for(k=0;k<2;k++)//运行两次
{
for(i=0;i<m;i++)//矩阵的行数
{
s=0;
for(j=0;j<n;j++)//矩阵的列数
{
cin>>a;
A[i][j]+=a;
if(k==1)//k=1时,第二次运行
{
l[j]+=abs(A[i][j]);
s=s+abs(A[i][j]);//计算一行的累加和
}
}
if((k==1)&&(s==0))
p++;
}
}
for(int t=0;t<n;t++)//统计列中的0列
{
if(l[t]==0)
q++;
}
// cout<<"行数"<<p<<endl;
// cout<<"列数"<<q<<endl;
cout<<p+q<<endl;
memset(A,0,sizeof(int)*100);
memset(l,0,sizeof(int)*10);
}
return 0;
}