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 MAX 100001
int Partition(int,int);
int A[MAX];
main(){
  int x,i,n;
  scanf("%d",&n);
  for(i = 0 ; i < n ; i++){
    scanf("%d",&A[i]);
  }
  x = Partition(0,n-1);
  for(i = 0  ; i < n ;i++){
    if(i==x) printf("[%d]",A[i]);
    else printf("%d",A[i]);   
    if(i==n-1) printf("\n");
    else printf(" ");
}
  return 0;
}

int Partition(int p,int r){
  int a,R=0,L=0,i,j;
  a=A[r];
  i=p-1;
  for(j = p ; j <= r-1 ; j++){
    if(A[j] <= a){ 
      i = i+1;
      R = A[i];
      A[i] = A[j];
      A[j] = R;
    }
  }
    L = A[i+1];
    A[i+1] = A[r];
    A[r] = L;
    return i+1;
}