Problem:
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Priority Queue

A priority queue is a data structure which maintains a set $S$ of elements, each of with an associated value (key), and supports the following operations:

$insert(S, k)$: insert an element $k$ into the set $S$

$extractMax(S)$: remove and return the element of $S$ with the largest key

Write a program which performs the $insert(S, k)$ and $extractMax(S)$ operations to a priority queue $S$.
The priority queue manages a set of integers, which are also keys for the priority.

Input

Multiple operations to the priority queue $S$ are given. Each operation is given by "insert $k$", "extract" or "end" in a line. Here, $k$ represents an integer element to be inserted to the priority queue.

The input ends with "end" operation.

Output

For each "extract" operation, print the element extracted from the priority queue $S$ in a line.

Constraints

The number of operations $\leq 2,000,000$

$0 \leq k \leq 2,000,000,000$

Sample Input 1

insert 8
insert 2
extract
insert 10
extract
insert 11
extract
extract
end

Sample Output 1

8
10
11
2

Solution in C (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#define N 2000000
#define NIL 2000000000
 
int A[N];
int hs = 0;
 
void maxHeap(int i){
  int l, r, lar;
  int temp;
  l = 2*i;
  r = 2*i+1;
  if(l <= hs && A[l] > A[i]) lar = l;
  else lar = i;
  if(r <= hs && A[r] > A[lar]) lar = r;
  if(lar != i){
    temp = A[i];
    A[i] = A[lar];
    A[lar] = temp;
    maxHeap(lar);
  }
}
void change(int i, int key){
  int temp;
  A[i] = key;
  while(i > 1 && A[i/2] < A[i]){
    temp = A[i];
    A[i] = A[i/2];
    A[i/2] = temp;
    i /= 2;
  }
}
 
void insert(int key){
  hs++;
  A[hs] = NIL;
  change(hs, key);
}
 
int extract(void){
  int max;
  max = A[1];
  A[1] = A[hs];
  hs--;
  maxHeap(1);
  return max;
}
 
int main(){
  int k;
  char word[10];
  while(1){
    scanf("%s", word);
    if(word[0] == 'i'){
      scanf("%d", &k);
      insert(k);
    }
    if(word[0] == 'e' && word[1] == 'x') printf("%d\n", extract());
    if(word[0] == 'e' && word[1] == 'n') break;
  }
return 0;
}