首页 > ACM题库 > 九度OJ > 九度-1001-A+B for Matrices
2013
12-13

九度-1001-A+B for Matrices

题目1001:A+B for Matrices

题目来源:2011年浙江大学计算机及软件工程研究生机试真题

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

简单的模拟题

#include <stdio.h>
#include <string.h>

#define M 12

int matrix[2][M][M];
int res[M + M];

int main(int argc, char **argv)
{
    int i, j, m, n, nresult;

    while (EOF != scanf("%d", &m) && m)
    {
        memset(res, 0, sizeof(res));
        nresult = 0;
        scanf("%d", &n);
        for (i = 0; i < m; ++i)
        {
            for (j = 0; j < n; ++j)
            {
                scanf("%d", &matrix[0][i][j]);
            }
        }
        for (i = 0; i < m; ++i)
        {
            for (j = 0; j < n; ++j)
            {
                scanf("%d", &matrix[1][i][j]);
                matrix[1][i][j] += matrix[0][i][j];
                if (matrix[1][i][j])
                {
                    res[i] = res[m + j] = 1;
                }
            }
        }
        for (i = 0; i < m + n; ++i)
        {
            if (!res[i])
            {
                ++nresult;
            }
        }
        printf("%d\n", nresult);
    }
    return 0;
}

 


  1. 这道题目的核心一句话是:取还是不取。
    如果当前取,则index+1作为参数。如果当前不取,则任用index作为参数。

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