sicily 1443 Printer Queue

1443. Printer Queue

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

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, thenmove 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 importantmuffin 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 simplifymatters, 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 linewith 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

31 054 21 2 3 46 01 1 9 1 1 1

Sample Output

125

这道题题意如下

有一个长度为n的打印任务队列，每个任务有优先级每次从队列头得到一个任务，如果它是剩余任务中优先级最高的，则打印它，否则放到队列尾 问其中某个任务是第几个被打印的。 n ≤100

我用list来模拟队列，注意，一定要在每次判断后更新iterator的值，不然会出现指针错误

// Problem#: 1443// Submission#: 3016334// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include <iostream>#include <string>#include <list>#include <algorithm>#include <cmath>#include <iomanip>#include <stack>#include <iterator>using namespace std;struct elem{    int index;    int num;};bool findBig(list<elem>& ls){    list<elem>::iterator it = ls.begin();    int first = (*it).num;    for ( ; it != ls.end(); it++)    {        if ((*it).num > first)            return true;    }    return false;}int main(){    int T;    cin >> T;    while (T--)    {        int n, m;        cin >> n >> m;        list<elem> q;        list<elem>::iterator it;        for (int i = 0; i < n; i++)        {            elem temp;            cin >> temp.num;            temp.index = i;            q.push_back(temp);        }        int res = 0;        it = q.begin();        int count = 0;        while (res == 0)        {            if (findBig(q)) // 若找到比第一个大的， 则把第一个元素加到队列末尾            {                it = q.begin();                elem temp = *it;                q.pop_front();                q.push_back(temp);                it = q.begin();            }            else if( (*it).index == m )// 如果找到了目标元素， 则把目标元素的count加到res，循环结束            {                res = count+1;                break;            }            else // 如果找到的不是目标元素，则打印并且count自增            {                q.pop_front();                count++;                it = q.begin();            }        }        cout << res << endl;    }    return 0;}