//題目1001：A+B for Matrices//時間限制：1 秒內存限制：32 兆特殊判題：否提交：20447解決：8174//題目描述：// 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.//輸入：// 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.//輸出：// For each test case you should output in one line the total number of zero rows and columns of A+B.//樣例輸入：//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//樣例輸出：//1//5#include "stdafx.h"#include "stdio.h"#include "string.h"#define MAX 11int a[MAX][MAX];int column[MAX];int row[MAX];int main(){ //freopen("1001_data.in","r",stdin); //freopen("1001_data.out","w",stdout); int n,m,temp,count; while(scanf("%d %d",&m,&n) == 2 && m!=0) { for(int i = 0;i<MAX;i++) { column[i]=1; row[i]=1; } count = 0; for(int i = 0;i<m;i++) for(int j=0; j<n ;j++) { scanf("%d",&a[i][j]); } for(int i=0;i<m;i++) { for(int j=0;j<n;j++) { scanf("%d",&temp); a[i][j] += temp; if(a[i][j]!=0) { row[i] = 0; column[j] = 0; } } count += row[i]; } for(int j = 0;j<n;j++) { count += column[j]; } printf("%d/n",count); } return 0;}
