【九度OJ】1001：A+B for Matrices

地址：
http://ac.jobdu.com/problem.php?pid=1001
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
來源：
2011年浙江大學計算機及軟件工程研究生機試真題

題目大意：
將兩個矩陣相加，求和矩陣中全是零的行和列的個數

源碼：

#include<stdio.h>

int n, m;
int matrixA[ 15 ][ 15 ];    //矩陣A，矩陣A+B
int matrixB[ 15 ][ 15 ];    //矩陣B

int main(){
    while( scanf( "%d", &m) != EOF ){
        if( m == 0 ){
            return 0;
        }
        scanf( "%d", &n );
        for( int i = 0; i < m; i ++ ){
            for( int j = 0; j < n; j ++ ){
                scanf( "%d", &matrixA[ i ][ j ] );
            }
        }
        for( int i = 0; i < m; i ++ ){
            for( int j = 0; j < n; j ++ ){
                scanf( "%d", &matrixB[ i ][ j ] );
            }
        }

        for( int i = 0; i < m; i ++ ){
            for( int j = 0; j < n; j ++ ){
                matrixA[ i ][ j ] += matrixB[ i ][ j ];
            }
        }

        int count = 0;
        //行
        for( int i = 0; i < m; i ++ ){
            int flag = 1;
            for( int j = 0; j < n; j ++ ){
                if( matrixA[ i ][ j ] != 0 ){
                    flag = 0;
                    break;
                }
            }
            if( flag == 1 ){
                count++;
            }
        }

        //列
        for( int j = 0; j < n; j ++ ){
            int flag = 1;
            for( int i = 0; i < m; i ++ ){
                if( matrixA[ i ][ j ] != 0 ){
                    flag = 0;
                    break;
                }
            }
            if( flag == 1 ){
                count++;
            }
        }

        printf( "%d\n", count );

    }
}