Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

Maximum Heap

A binary heap which satisfies max-heap property is called max-heap. In a max-heap, for every node $i$ other than the root, $A[i] \leq A[parent(i)]$, that is, the value of a node is at most the value of its parent. The largest element in a max-heap is stored at the root, and the subtree rooted at a node contains values no larger than that contained at the node itself.

Here is an example of a max-heap.

Write a program which reads an array and constructs a max-heap from the array based on the following pseudo code.

$maxHeapify(A, i)$ move the value of $A[i]$ down to leaves to make a sub-tree of node $i$ a max-heap. Here, $H$ is the size of the heap.

1 maxHeapify(A, i)
2 l = left(i)
3 r = right(i)
4 // select the node which has the maximum value
5 if l ≤ H and A[l] > A[i]
6 largest = l
7 else
8 largest = i
9 if r ≤ H and A[r] > A[largest]
10 largest = r
11
12 if largest ≠ i　// value of children is larger than that of i
13 swap A[i] and A[largest]
14 maxHeapify(A, largest) // call recursively

The following procedure buildMaxHeap(A) makes $A$ a max-heap by performing maxHeapify in a bottom-up manner.

1 buildMaxHeap(A)
2 for i = H/2 downto 1
3 maxHeapify(A, i)

Input

In the first line, an integer $H$ is given. In the second line, $H$ integers which represent elements in the binary heap are given in order of node id (from $1$ to $H$).

Output

Print values of nodes in the max-heap in order of their id (from $1$ to $H$). Print a single space character before each value.

Constraint

$1 \leq H \leq 500,000$

$-2,000,000,000 \leq$ value of a node $\leq 2,000,000,000$

Sample Input 1

10
4 1 3 2 16 9 10 14 8 7

Sample Output 1

16 14 10 8 7 9 3 2 4 1

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in C (judge verdict: Accepted)

#include<stdio.h>

/* 値を入れ替える関数 */
void swap (int *x, int *y) {
  int temp;    // 値を一時保存する変数

  temp = *x;
  *x = *y;
  *y = temp;
}

/* pushdouwn操作 */
void pushdown (int array[], int first, int last) {
  int parent = first;   // 親
  int child = 2*parent;   // 左の子
  while (child <= last) {
    if ((child < last) && (array[child] < array[child+1])) {
       child++;   // 右の子の方が大きいとき、右の子を比較対象に設定
    }
    if (array[child] <= array[parent]) { break; }   // ヒープ済み
    swap(&array[child], &array[parent]);
    parent = child;
    child = 2* parent;
  }
}


int main (void) {
  int array[500001];
  int i,n;

  scanf("%d",&n);

  for(i=1;i<=n;i++){
    scanf("%d",&array[i]);
      }

  for (i = n/2; i >= 1; i--) {
    pushdown(array, i, n);   // 全体をヒープ化
  }

  for(i=1;i<=n;i++){
    printf(" %d",array[i]);
  }

  printf("\n");
    


  return 0;
}