题目链接:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3252
Time limit 3000 ms

Description

The only printer in the computer science students’
union is experiencing an extremely heavy workload.
Sometimes there are a hundred jobs in the printer queue and you may have to wait for hours to get a single page of output.
Because some jobs are more important than others, the Hacker General has invented and implemented a simple priority system for the print job queue. Now, each job is assigned a priority between 1 and 9 (with 9 being the highest priority, and 1 being the lowest), and the printer operates as follows.

• The first job J in queue is taken from the queue.
• If there is some job in the queue with a higher priority than job J, then move J to the end of the queue without printing it.
• Otherwise, print job J (and do not put it back in the queue).
In this way, all those important muffin recipes that the Hacker General is printing get printed very quickly. Of course, those annoying term papers that others are printing may have to wait for quite some time to get printed, but that’s life.
Your problem with the new policy is that it has become quite tricky to determine when your print job will actually be completed. You decide to write a program to figure this out. The program will be given the current queue (as a list of priorities) as well as the position of your job in the queue, and must then calculate how long it will take until your job is printed, assuming that no additional jobs will be added to the queue. To simplify matters, we assume that printing a job always takes exactly one minute, and that adding and removing jobs from the queue is instantaneous.

Input

One line with a positive integer: the number of test cases (at most 100). Then for each test case:
• One line with two integers n and m, where n is the number of jobs in the queue (1 ≤ n ≤ 100)
and m is the position of your job (0 ≤ m ≤ n − 1). The first position in the queue is number 0, the second is number 1, and so on.
• One line with n integers in the range 1 to 9, giving the priorities of the jobs in the queue. The first integer gives the priority of the first job, the second integer the priority of the second job, and so on.

Output

For each test case, print one line with a single integer; the number of minutes until your job is completely
printed, assuming that no additional print jobs will arrive.

Sample Input

3
1 0
5
4 2
1 2 3 4
6 0
1 1 9 1 1 1

Problem solving report:

Description: 一个打印序列，有优先级别顺序。从前往后扫描，若该作业优先级别已是最高，则执行此作业。若不是，放到队列末尾。求指定位置的作业W被执行的时间
Problem solving: 直接模拟，先执行所有比作业m优先级高的作业，当然也要按照优先级从高到低顺序执行。从队列首开始扫描，到达指定作业m，停止扫描。 

Accepted Code:

#include <bits/stdc++.h>
using namespace std;
int main() {
    int t, n, m, a;
    scanf("%d", &t);
    while (t--) {
        queue <int> Q;
        priority_queue <int> S;
        scanf("%d%d", &n, &m);
        for (int i = 0; i < n; i++) {
            scanf("%d", &a);
            Q.push(a), S.push(a);
        }
        while (true) {
            int u = Q.front();
            Q.pop();
            if (!m){//m=0,则证明现在打印的是目标任务
                if (u != S.top()) {//队列中还有优先级比x高的
                    m = S.size() - 1;//放到队尾
                    Q.push(u);//将该任务放到队尾
                }
                else break;
            }
            else {//现在的任务还不是目标任务
                m--;//前面的任务减1 
                if (u != S.top())
                    Q.push(u);
                else S.pop();
            }
        }
        printf("%d\n", n - Q.size());//Q.size是全部打印结束后Q队列的长度，即还未打印的数目 
    }
    return 0;

}