There is a bag-like data structure, supporting two operations:
1 x Throw an element x into the bag. 2 Take out an element from the bag.
Given a sequence of operations with return values, you’re going to guess the data structure. It is a stack (Last-In, First-Out), a queue (First-In, First-Out), a priority-queue (Always take out larger elements first) or something else that you can hardly imagine!
Input There are several test cases. Each test case begins with a line containing a single integer n (1 ≤ n ≤ 1000). Each of the next n lines is either a type-1 command, or an integer 2 followed by an integer x. That means after executing a type-2 command, we get an element x without error. The value of x is always a positive integer not larger than 100. The input is terminated by end-of-file (EOF).

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Output
For each test case, output one of the following:
stack It’s definitely a stack. queue It’s definitely a queue. priority queue It’s definitely a priority queue. impossible It can’t be a stack, a queue or a priority queue. not sure It can be more than one of the three data structures mentioned above.

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Sample Input

6

1 1     1 2     1 3     2 1     2 2     2 3

6

1 1     1 2      1 3     2 3     2 2     2 1

2

1 1     2 2

4

1 2     1 1      2 1      2 2

7

1 2     1 5      1 1      1 3     2 5      1 4       2 4

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Sample Output
queue

not sure

impossible

stack

priority queue

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题意：根据操作，判断属于那种数据结构。

思路：使用STL中对应函数，进行模拟，若矛盾则不符合。

#include<stdio.h>
#include<queue>
#include<stack>
using namespace std;
int main()
{
	int n;
	while (~scanf("%d", &n))
	{
		stack<int>st;
		queue<int>qu;
		priority_queue<int>pr;
		int a = 1, b = 1, c = 1;
		for (int i = 0; i < n; i++)
		{
			int x, y;
			scanf("%d %d", &x, &y);
			if (x == 1)//向其中加入数据
			{
				st.push(y);
				qu.push(y);
				pr.push(y);
			}
			else
			{
				if (st.empty())
				{
					a = b = c = 0;
					continue;
				}
				if (a)
					a = (st.top() == y);//判断顶部的数是否与y相等
				if (b)
					b = (qu.front() == y);
				if (c)
					c = (pr.top() == y);
				st.pop();
				qu.pop();
				pr.pop();
			}
		}
		if (a + b + c > 1)
			printf("not sure\n");
		else if (a)
			printf("stack\n");
		else if (b)
			printf("queue\n");
		else if (c)
			printf("priority queue\n");
		else
			printf("impossible\n");
	}

	return 0;
}