时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：13653

解决：5575

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

题意：给出2个n*m的矩阵，然后相加，判断有多少行或者列的值均为0

import java.util.Scanner;
public class Main 
{
    public static void main(String[] args) 
    {
        Scanner cin = new Scanner(System.in);
        int [][]a = new int[20][20];
        int n,m;
        while(1>0)
        {
            n=cin.nextInt();
            if(n==0)return ;
            m=cin.nextInt();
            for (int i = 0; i<n; i++)
            {
                for (int j = 0; j < m; j++)
                {
                    a[i][j] = cin.nextInt();
                }
            }
            for (int i = 0; i<n; i++)
            {
                for (int j = 0; j < m; j++)
                {
                    int t = cin.nextInt();
                    a[i][j] +=t; 
                }
            }
            int count = 0;
            for(int i=0,j=0;i<n;i++)
            {
                for(j=0;j<m;j++)
                {
                    if(a[i][j]!=0)
                        break;
                }
                if(j==m) count++;
            }
            for(int i=0,j=0;i<m;i++)
            {
                for(j=0;j<n;j++)
                {
                    if(a[j][i]!=0)
                        break;
                }
                if(j==n) count++;
            }
            System.out.println(count);
        } 
    }  
}