1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 class minStack{
 5 private:
 6     stack<int> s;
 7     stack<int> mins;
 8 public:
 9     void push(int x){
10         s.push(x);
11         if(mins.empty()==true || mins.top()>=x)
12             mins.push(x);
13     }
14     
15     int getMin(){
16         return mins.top();
17     }
18     
19     void pop(){
20         int temp = s.top();
21         s.pop();
22         if(temp == mins.top())
23             mins.pop();
24     }
25 };
26 
27 int main(){
28     int N;
29     cin>>N;
30     minStack ms;
31     for(int i=0;i<N;++i){
32         string s = "";
33         cin>>s;
34         if(s == "push"){
35             int num;
36             cin>>num;
37             ms.push(num);
38         }
39         else if(s == "pop"){
40             ms.pop();
41         }
42         else
43             cout<<ms.getMin()<<endl;
44     }
45     return 0;
46 }

最小栈问题，众所周知系列，就是用另一个min栈来存放当前栈中的最小元素。问题不大