九度oj 1001 A+B for Matrices 2011年浙江大学计算机及软件工程研究生机试真题

题目1001：A+B for Matrices

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：15235

解决：6172

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

来源：

2011年浙江大学计算机及软件工程研究生机试真题

分析：

读懂题意就OK

 #include<iostream>
 #include<queue>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 using namespace std;
 int map[][];
 int main(){
     int m,n;
     while(scanf("%d",&m)&&m){
     scanf("%d",&n);
     int i,j;
     for(i=;i<m;i++){
         for(j=;j<n;j++){
             scanf("%d",&map[i][j]);
         }
     }
     for(i=;i<m;i++){
         for(j=;j<n;j++){
             int t;
             scanf("%d",&t);
             map[i][j]+=t;
     }
     }
     int sum=;
     for(i=;i<m;i++){
         for(j=;j<n;j++){
             if(map[i][j])
             break;
         }
         if(j==n)
         sum++;
     }
     for(i=;i<n;i++){
         for(j=;j<m;j++){
             if(map[j][i])
             break;
         }
         if(j==m)
         sum++;
     }
     cout<<sum<<endl;
     }
     return ;
 }