Problem:
Partition

Quick sort is based on the Divide-and-conquer approach. In QuickSort(A, p, r), first, a procedure Partition(A, p, r) divides an array A[p..r] into two subarrays A[p..q-1] and A[q+1..r] such that each element of A[p..q-1] is less than or equal to A[q], which is, inturn, less than or equal to each element of A[q+1..r]. It also computes the index q.

In the conquer processes, the two subarrays A[p..q-1] and A[q+1..r] are sorted by recursive calls of QuickSort(A, p, q-1) and QuickSort(A, q+1, r).

Your task is to read a sequence A and perform the Partition based on the following pseudocode:

Partition(A, p, r)
1 x = A[r]
2 i = p-1
3 for j = p to r-1
4 do if A[j] <= x
5 then i = i+1
6 exchange A[i] and A[j]
7 exchange A[i+1] and A[r]
8 return i+1

Note that, in this algorithm, Partition always selects an element A[r] as a pivot element around which to partition the array A[p..r].

Input

The first line of the input includes an integer n, the number of elements in the sequence A.

In the second line, Ai (i = 1,2,...,n), elements of the sequence are given separated by space characters.

Output

Print the sorted sequence. Two contiguous elements of the sequence should be separated by a space character. The element which is selected as the pivot of the partition should be indicated by [  ].

Constraints

1 ≤ n ≤ 100,000

0 ≤ Ai ≤ 100,000

Sample Input 1

12
13 19 9 5 12 8 7 4 21 2 6 11

Sample Output 1

9 5 8 7 4 2 6 [11] 21 13 19 12

Solution in C (judge verdict: Accepted)

#include <stdio.h>
#define N 100000

int partition(int *,int ,int );
int A[N];

int main()
{
	int i,j;
	int n,q;
	
	scanf("%d",&n);
	for(i=0;i<n;i++){
		scanf("%d",&A[i]);
	}
	q = partition(A,0,n-1);
	for(i=0;i<n;i++){
		if(i==q){
			printf("[%d] ",A[i]);
		} 
		else if(i != n-1) {
			printf("%d ",A[i]);
		}
		else {
			printf("%d",A[i]);
		}
	}
	printf("\n");
	
	return 0;
}

int partition(int *A,int p ,int r){
	int tmp;
	int i,j,x;
	
	x = A[r];
	i = p-1;
	
	for(j=0;j<r;j++){
		if(A[j] <= x){
			i++;
			tmp = A[j];
			A[j] = A[i];
			A[i] = tmp;
		}
	}
	tmp = A[r];
	A[r] = A[i+1];
	A[i+1] = tmp;
	
	return i+1;
}