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：21870

解决：8649

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 21 11 1-1 -110 92 31 2 34 5 6-1 -2 -3-4 -5 -60

样例输出：

1
5

#include<stdio.h>#include<iostream>using namespace std;const int Max=10;int main(){int M,N;int A[Max][Max],B[Max][Max];int tmp;while(1){scanf("%d%d",&M,&N);if(M==0)break;if(M<0||M>10||N<0||N>10)return false;for(int i=0;i<M;i++){for(int j=0;j<N;j++){scanf("%d",&tmp);if(tmp<-100||tmp>100)return false;A[i][j]=tmp;}}for(int i=0;i<M;i++){for(int j=0;j<N;j++){scanf("%d",&tmp);if(tmp<-100||tmp>100)return false;B[i][j]=tmp;}}for(int i=0;i<M;i++){for(int j=0;j<N;j++){A[i][j]=A[i][j]+B[i][j];}}int count=0;for(int i=0;i<M;i++){    int tag=1;for(int j=0;j<N;j++){if(A[i][j]!=0){tag=0;break;}}if(tag==1)count++;}for(int j=0;j<N;j++){    int tag=1;for(int i=0;i<M;i++){if(A[i][j]!=0){tag=0;break;}}if(tag==1)count++;}printf("%d\n",count);}return 0;}