九度oj 题目1001：A+B for Matrices 【ZJU2011考研机试题1】

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：11539

解决：4694

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

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对二维数组的遍历

/**************************************************************
Problem: 1001
Language: C++
Result: Accepted
Time:0 ms
Memory:1080 kb
****************************************************************/
#include <stdio.h>
#include<string.h>
const int MAX=1000;
int ans[MAX][15],cnt;
int main()
{
int n,m,tmp,i,j;
//freopen("G:\\in.txt", "r", stdin);
while(scanf("%d%d",&n,&m)!=EOF){
if(n==0) break;
for(i=0;i<n;i++){  //输入n行作为矩阵A
for(j=0;j<m;j++)
scanf("%d",&ans[i][j]);
}
for(i=0;i<n;i++){  //输入n行作为矩阵A+B
for(j=0;j<m;j++){
scanf("%d",&tmp);
ans[i][j]+=tmp;
}
}
cnt=0;
for(i=0;i<n;i++){
for(j=0;j<m;j++){
if(ans[i][j]!=0)
break;
}
if(j==m)
cnt++;
}
for(i=0;i<m;i++){
for(j=0;j<n;j++){
if(ans[j][i]!=0)
break;
}
if(j==n)
cnt++;
}
printf("%d\n",cnt);
}
return 0;
}