Author:  Andrew Stankevich (original idea, text, solution)  Time limit:  2 sec  
Input file:  heapsort.in  Memory limit:  64 Mb  
Output file:  heapsort.out 
A well known algorithm called heapsort is a deterministic sorting algorithm taking O(n log n) time and O(1) additional memory. Let us describe ascending sorting of an array of different integer numbers.
The algorithm consists of two phases. In the first phase, called heapification, the array of integers to be sorted is converted to a heap. An array a[1…n] of integers is called a heap if for all 1 ≤ i ≤ n the following heap conditions are satisfied:
 if 2i ≤ n then a[i] > a[2i];
 if 2i + 1 ≤ n then a[i] > a[2i + 1].
We can interpret an array as a binary tree, considering children of element a[i] to be a[2i] and a[2i + 1]. In this case the parent of a[i] is a[i div 2], where i div 2 = floor(i / 2). In terms of trees the property of being a heap means that for each node its value is greater than the values of its children.
In the second phase the heap is turned into a sorted array. Because of the heap condition the greatest element in the heapified array is a[1]. Let us exchange it with a[n], now the greatest element of the array is at its correct position in the sorted array. This is called extractmax.
Now let us consider the part of the array a[1 ... n1]. It may be not a heap because the heap condition may fail for i=1. If it is so (that is, either a[2] or a[3], or both are greater than a[1]) let us exchange the greatest child of a[1] with it, restoring the heap condition for i=1. Now it is possible that the heap condition fails for the position that now contains the former value of a[1]. Apply the same procedure to it, exchanging it with its greatest child. Proceeding so we convert the whole array a[1 ... n1] to a heap. This procedure is called sifting down. After converting the part a[1 ... n1] to a heap by sifting, we apply extractmax again, putting second greatest element of the array to a[n1], and so on.
For example, let us see how the heap a=(5, 4, 2, 1, 3) is converted to a sorted array. Let us make the first extractmax. After that the array turns to (3, 4, 2, 1, 5). Heap condition fails for a[1] = 3 because its child a[2] = 4 is greater than it. Let us sift it down, exchanging a[1] and a[2]. Now the array is (4, 3, 2, 1, 5). The heap condition is satisfied for all elements, so sifting is over. Let us make extractmax again. Now the array turns to (1, 3, 2, 4, 5). Again the heap condition fails for a[1]; exchanging it with its greatest child we get the array (3, 1, 2, 4, 5) which is the correct heap. So we make extractmax and get (2, 1, 3, 4, 5). This time the heap condition is satisfied for all elements, so we make extractmax, getting (1, 2, 3, 4, 5). The leading part of the array is a heap, and the last extractmax finally gives (1, 2, 3, 4, 5).
It is known that heapification can be done in O(n) time. Therefore, the most time consuming operation in heapsort algorithm is sifting, which takes O(n * log (n)) time.
In this problem you have to find a heapified array containing different numbers from 1 to n, such that when converting it to a sorted array, the total number of exchanges in all sifting operations is maximal possible. In the example above the number of exchanges is 1+1+0+0+0 = 2, which is not the maximum. (5, 4, 3, 2, 1) gives the maximal number of 4 exchanges for n=5.
No.  Input file (heapsort.in ) 
Output file (heapsort.out ) 

1 

