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>

int S[2000000], n = 0;

void maxHeapify(int i){
  int l, r, largest, a;

  l = i * 2;
  r = i * 2 + 1;

  if(l <= n && S[l] > S[i])
    largest = l;
  else
    largest = i;

  if(r <= n && S[r] > S[largest])
    largest = r;

  if(largest != i){
    a = S[i];
    S[i] =  S[largest];
    S[largest] = a;
    maxHeapify(largest);
  }
}
int extractMax(){
  int i, e;
  if(n < 1) return -10000000;
  e = S[1];
  S[1] = S[n--];
  maxHeapify(1);
  return e;
}

void increk(int i, int k){
  int a;
  if(k < S[n]) return;
  S[n] = k;
  while(i > 1 && S[i / 2] < S[i]){
    a = S[i];
    S[i] = S[i / 2];
    S[i / 2] = a;
    i = i / 2;
  }
}

void insert(int k){
  n++;
  S[n] = -10000000;
  increk(n, k);
}

int main(){
  int key;
  char a[10];

  while(1){
    scanf("%s", a);
    if(a[0] == 'e' && a[1] == 'n') break;
    if(a[0] == 'i') {
      scanf("%d", &key);
      insert(key);
    }
    else printf("%d\n", extractMax());
  }

  return 0;
}