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>

#define SIZE 2000000

int main() {
  int A[SIZE];
  int H=0;
  int i, l, largest, tmp, n;
  char com[21], buf[20];
  buf[19]='\0';
  char *p;
  do {
    fgets(com, 20, stdin);
    if (com[0] == 'i') {
      n = 0;
      p = com+7;
      while (*p!='\n') {
        n = n*10+*p-'0';
        p++;
      }
      A[++H] = n; //myatoi(com+7);
      for (i=H; i>1 && A[i/2]<A[i];i/=2) {
        tmp = A[i/2];
        A[i/2] = A[i];
        A[i] = tmp;
      }

    } else if (com[1]=='x') { // extract
      n = A[1];
      for (i=18; n>0; n/=10) buf[i--]=(n%10)+'0';
      puts(buf+i+1);
      A[1] = A[H];
      i = 1;
      while (1) {
        l = i*2;
        largest = i;
        if (l < H && A[l] > A[i])        largest = l;
        if (l+1 < H && A[l+1] > A[largest])  largest = l+1;
        if (largest == i) break;
        tmp = A[i];
        A[i] = A[largest];
        A[largest] = tmp;
        i = largest;
      }
      H--;
    }
  } while (com[2]!='d');
  return 0;
}