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>

void  swap(int *x, int *y)
{
    int  temp;
    temp = *x;        /* xの値を一時保存 */
    *x = *y;          /* xの値をyの値に書き換える */
    *y = temp;        /* yの値をxの値に書き換える */
}

int Partition(int A[],int p,int r){
	//基準値をxに記録。
	int x = A[r];
	int i = p;
	
	/*
	pからiまで<=x, i+1からjまでをx<。
	jを増やしながら基準値とA[j]を比べる。とにかく左側に小さいのを持ってくる。
	A[j]の方が大きい場合は何もしない。
	A[j]の方が小さい場合はA[j]をA[i]に持ってってiを１増やす。
	*/
	for(int j = p; j < r; j++){
		if(A[j] <= x){
			swap(&A[i],&A[j]);
			i++;
		}
	}
	swap(&A[i],&A[r]); //このswapはfor文の中に組み込んでもいい。その場合はj<=rでreturn i-1。
	return i;
}

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