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九度OJ-1001-A+B矩阵-有些小技巧

2024-10-14 14:05:02   212人阅读

题目1001：A+B for Matrices

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：22974

解决：9099

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

 

 

 

 

 

#include <stdio.h>
int ma[11][11];
int mb[11][11];

int main()    {
    int a, b;
    while(scanf("%d", &a))    {
        
        if(a == 0)    return 0;
        scanf("%d", &b);
        
        for(int i = 0; i < a; i++)    {
            for(int j = 0; j < b; j++)    {
                scanf("%d", &ma[i][j]);
            }
        }
        for(int i = 0; i < a; i++)    {
            for(int j = 0; j < b; j++)    {
                scanf("%d", &mb[i][j]);
                mb[i][j] += ma[i][j];
            }
        }
        
        // sum
        int su = 0;
        int co = 0;
        for(int i = 0; i < a; i++)    {
            for(int j = 0; j < b; j++)    {
                if(mb[i][j] != 0){
                    co++;
                    // printf("%d %d\n", i, j);
                    break;
                }    
                
            }
            if(co == 0)    {
                su++;
                // printf("%d %d\n", i, su);
            }
            else co = 0;
        }
        
        co = 0;
        for(int j = 0; j < b; j++)    {
            for(int i = 0; i < a; i++)    {
                if(mb[i][j] != 0)    {
                    co++;
                    break;
                }    
                
            }
            if(co == 0)    {
                su++;
            }
            else co = 0;
        }
        
        printf("%d\n", su);
    }
    return 0;
}

 

九度OJ-1001-A+B矩阵-有些小技巧


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