题目
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 ?rst 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 mu?n 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 ?gure 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 ?rst 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 ?rst integer gives the priority of the ?rst 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.
思路
解这道题目的关键是要在常数时间内获取当前队列中最大的优先级，还有怎样记录自己文档的位置；
因为优先级只有1到9，所以这里我使用了桶排方法记录每个优先级的出现次数，然后每次从9到1寻找值大于0的第一个优先级，这个就是目前队列中存在的最大的优先级。
然后对于自己的文档，由于给出位置m，所以在第m次读入数据时，我就将自己文档的优先级储存为负数，这样就不用去考虑自己文档在队列中的位置了，只要出队的数据为负数就可以结束操作，输出结果。
代码
#include <iostream>
#include <cmath>
#include <queue>
using namespace std;
int getPro(int a[])
{
for(int i = 9; i >= 0; i--)if(a[i] > 0) return i;
}
int main()
{
// freopen("i.txt", "r", stdin);// freopen("o.txt", "w", stdout);int k, n, m, num; scanf("%d", &k);while(k--){
scanf("%d%d", &n, &m);queue<int> q;int a[10] = {
0}; a[0] = 1;for(int i = 0; i < n; i++){
scanf("%d", &num);a[num]++;if(i == m) num = -num;q.push(num);}int max_pro = getPro(a), ans = 0;while(!q.empty()){
int fir = q.front(); q.pop();if(abs(fir) < max_pro) q.push(fir);else{
if(fir < 0){
printf("%d\n", ++ans); break;}else{
ans++; a[fir]--;max_pro = getPro(a);}}}}return 0;
}