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>
#include <stdlib.h>
#include <iostream>
#include <stack>
#define N_MAX 100000
using namespace std;
//常に心してかかれ
//簡単なことを簡単にできるように練習する
//びっくりするくらいつまらないことで詰まってることも多い

int n;

int partition(int A[],int p,int r){//rは配列の最後を指す添え字
	int x = A[r];
	int i = p;
	int tmp;
	
	for(int j=p;j<r;j++){
		if(A[j] <= x){
			tmp = A[i];
			A[i] = A[j];
			A[j] = tmp;
			i++;
		}
	}
	tmp = A[i];
	A[i] = A[r];
	A[r] = tmp;

	return i;//iはx以下の数の数ともいえる
}

int main(){
	int A[N_MAX+1];
	int p;

	scanf("%d",&n);

	//確認用
	//printf("%d\n",n);
	
	for(int i=0;i<n;i++){
		scanf("%d",&A[i]);

		//確認用
		//printf("%d ",A[i]);
	}

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