1 条题解

  • 0
    @ 2024-10-19 19:36:13

    看懂题意其实就很好做了

    意思就是没有完成某个依赖性的任务之前,后面的任务是没有办法完成的(即便完成了也得不到分数)

    所以我们只要模拟一下。如果某一题没做出来,会影响到什么题目。

    #include <bits/stdc++.h>
    using namespace std;
    int a[5][5];
    int score[5];
    int main()
    {
        freopen("chen.in", "r", stdin);
        freopen("chen.out", "w", stdout);
        for (int i = 1; i <= 4; i++)
            for (int j = 1; j <= 4; j++)
                cin >> a[i][j];
        for (int no = 1; no <= 4; no++)
        {
            for (int i = 1; i <= 4; i++)
                score[i] = i * 10;
            score[no] = 0;
            if (score[1] == 0 && a[1][2] == 1)
                score[2] = 0;
            if (score[1] == 0 && a[1][3] == 1)
                score[3] = 0;
            if (score[1] == 0 && a[1][4] == 1)
                score[4] = 0;
    
            if (score[2] == 0 && a[2][3] == 1)
                score[3] = 0;
            if (score[2] == 0 && a[2][4] == 1)
                score[4] = 0;
    
            if (score[3] == 0 && a[3][4] == 1)
                score[4] = 0;
    
            cout << score[1] + score[2] + score[3] + score[4] << " ";
        }
        return 0;
    }
    
    • 1

    信息

    ID
    1038
    时间
    1000ms
    内存
    256MiB
    难度
    9
    标签
    递交数
    13
    已通过
    4
    上传者