单调栈运用2----(xdoj-1156)

#include <bits/stdc++.h>
using namespace std;
const int N=1e5+7;
int num[N];
int len;
int top;
int t;
deque <int> q;
int main ()
{
  int n;
  while (~scanf ("%d",&n) )           {
    while (!q.empty()) q.pop_back();
    len=t=0;
    top=1;
    while (n--) {
      int op,x;
      scanf ("%d",&op);
      if (op==1) {
        scanf ("%d",&x); x-=t;
        num[++len]=x;
        while (!q.empty()&&num[q.back()]<=x) q.pop_back();
        q.push_back(len);
      }
      else if (op==2) { // 想一想为什么要加括号 ？否则1句和2句就要连接在一块了
        if ( (top++)==q.front() )  q.pop_front(); //1
      }  
      else                                        // 2
        printf ("%d
",num[q.front()]+t); 
      t++;
    }
  }
  return 0;
}