题解 | #A+B for Matrices#

#include <iostream>
#include <cstring> 
#include <algorithm>
#include <math.h>
#include <stdlib.h>
using namespace std;
#define M 10
#define N 10
/*
题意：
	输入：第一行输入矩阵的行列数；然后按顺序输入A、B的元素值；
	输出：A+B之后，可能会有等于0的行或列，输出等于0的 行+列的总数
*/
int main(){
	int m,n;
	while(cin >>m>>n && m != 0 && n != 0){
		int A[M][N],B[M][N];
		//输入A
		for (int i = 0; i < m; i++){
			for (int j = 0; j < n; j++){
				cin >>A[i][j];
			}
		}

		//输入B
		for (int i = 0; i < m; i++){
			for (int j = 0; j < n; j++){
				cin >>B[i][j];

			}
		}
		//A+B
		for (int i = 0; i < m; i++){
			for (int j = 0; j < n; j++){
				A[i][j] +=B[i][j];
			}
		}

		
		int count = 0;
		//计算等于0的行数
		for (int i = 0; i < m; i++){
			int flag = 1;
			for (int j = 0; j < n; j++){
				if (A[i][j] != 0) flag = 0;
			}
			if (flag) count++;
		}

		//计算等于0的列数
		for (int i = 0; i < n; i++){
			int flag = 1;
			for (int j = 0; j < m; j++){
				if (A[j][i] != 0) flag = 0;
			}
			if (flag) count++;
		}
		cout<<count<<endl;
	}
}