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 n;
int A[N];
int Partition(int, int);
int main(){
	int i;
	int p=0;
	int r;
	int ans;
	scanf("%d", &n);
	for( i=0; i<n; i++ ){
		scanf("%d", &A[i]);
	}
	r=n-1;
	ans=Partition(p, r);
	for( i=0; i<ans; i++ ) printf("%d ", A[i]);
	printf("[%d] ", A[ans]);
	for( i=ans+1; i<n; i++ ){
		printf("%d", A[i]);
		if( i+1 < n) printf(" ");
	}
	printf("\n");
	return 0;
}

int Partition(int p, int r){
	int i;
	int j;
	int tmp;
	int x;
	x=A[r];
	i=p-1;
	for( j=p; j<=r-1; 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;
}