1.题目描述：点击打开链接
2.解题思路：本题要求根据输入的数据和输出的数据来猜测一种可能的数据结构，备选答案有“栈，队列，优先队列”，结果也可能都不是或者不确定。STL中已经有这三种数据结构了，因此直接模拟题意，输出时判断是否对应即可。注意：弹出时要判断一下是否已经为空。
3.代码：
#define _CRT_SECURE_NO_WARNINGS #include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<functional>using namespace std;stack<int>s;queue<int>q;priority_queue<int>p;char st[][20] = { "impossible", "priority queue", "queue","not sure" , "stack", "not sure","not sure" ,"not sure" };void clear(){while (!s.empty())s.pop();while (!q.empty())q.pop();while (!p.empty())p.pop();}void add(int x){s.push(x);q.push(x);p.push(x);}void pop(){if (!s.empty())s.pop();if (!q.empty())q.pop();if (!p.empty())p.pop();}int main(){//freopen("t.txt", "r", stdin);int n;while (~scanf("%d", &n)){clear();int ok1, ok2, ok3;ok1 = ok2 = ok3 = 1;for (int i = 0; i < n; i++){int a, b;cin >> a >> b;if (a == 1)add(b);//统一添加else{if (s.empty() || s.top() != b)ok1 = 0;if (q.empty() || q.front() != b)ok2 = 0;if (p.empty() || p.top() != b)ok3 = 0;pop();//统一弹出}}int x = (ok1 << 2) | (ok2 << 1) | ok3;//编码，方便输出结果printf("%s\n", st[x]);}return 0;}
，呼唤你前往另一个地方，过上另一种生活。
