九度OJ-1001

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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题目的意思大概是输入两个数作为数组的行数和列数，然后读入两个这样的数组，执行加法操作，算出元素均为0的行数列数的和。

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AC代码：

#include<iostream>using namespace std;int main(){    int m,n; //读入数组的行列数     int i,j;    int count1,count2=0; //计数器     while(cin>>m>>n&&m){        int a[m][n];        int b[m][n];        count2=0; //归零操作         for(i=0;i<m;i++)  //读入第一个数组            for(j=0;j<n;j++)                cin>>a[i][j];        for(i=0;i<m;i++)  //读入第二个数组             for(j=0;j<n;j++){                 cin>>b[i][j];                a[i][j]+=b[i][j]; //执行数组的加法            }         for(i=0;i<m;i++){  //算出有多少行是元素均为0的             for(j=0;j<n;j++)                        if(a[i][j]==0)                    count1++;                if(count1==n)                    count2++;                count1=0;  //归零操作             }        for(j=0;j<n;j++){ //算出有多少列是元素均为0的             for(i=0;i<m;i++)                if(a[i][j]==0)                    count1++;                if(count1==m)                    count2++;                count1=0; //归零操作             }        cout<<count2<<endl;     }    return 0;            } 

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