pat甲级1014

5个小时的debug体验，很酸了（一开始把m和n弄混，后来又发现只要在540之前接受服务就可以算成功。解法十分的粗暴，一分钟判断一次的逻辑.....代码不算复杂，细节比较恼人，不适合我这种无脑马大哈

#include <iostream>
#include <vector>
#include <queue>

using namespace std;

int main(void)
{
    //queue<int> wait;

    int n, m, k, q;
    cin >> n >> m >> k >> q;
    vector<int> cost_time(k + 2);//end
    vector<int> beg_time(k + 2, 541);//开始被服务
    for (int i = 0;i < k;i++)
        cin >> cost_time[i];
    vector<queue<int> > win(n);
    for (int i = 0;i < n;i++)//第一批入黄线
    {
        if(i<k)//记得加判断，有一个测试点n>k
            beg_time[i] = 0;
        for (int j = 0;j < m;j++)
        {
            if (n*j + i >= k)break;
            win[i].push(n*j + i);
        }
    }

    
    int next = n*m;//下一个入队，也有k<n*m的情况，不过下面做了判断
    for (int times = 1;times<540;times++)
    {
        for (int w = 0;w < n;w++)
        {
            if (win[w].empty())continue;
            int cus = win[w].front();
            if (cost_time[cus] == times&&beg_time[cus] < 540)
            {
                win[w].pop();
    
                if (next < k)win[w].push(next++);
                if (!win[w].empty())
                {
                    cost_time[win[w].front()] += cost_time[cus];
                    beg_time[win[w].front()] = cost_time[cus];
                }
            }
        }
    }
    //give ans
    for (int i = 0;i < q;i++)
    {
        int query;
        cin >> query;
        if (beg_time[query - 1] >= 540)
        {
            printf("Sorry
");
            continue;
        }

        int h = 8 + cost_time[query - 1] / 60, m = cost_time[query - 1] % 60;
        printf("%02d:%02d
", h, m);
    }

    return 0;
}