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 A[N];
int x;

int main() {
  int i,res;

  scanf("%d",&x);
  for(i = 0; i < x; i++) {
    scanf("%d",&A[i]);
  }

  res = partition(0,x - 1);

  for(i = 0; i < x; i++) {
    if(i) {
      printf(" ");
    }
    if(i == res) {
      printf("[");
    }
    printf("%d",A[i]);
   
    if(i == res) {
      printf("]");
    }
    
  }
  printf("\n");
  return 0;
}

int partition(int a, int b) {
  int y,i,j,t;

  y = A[b];
  i = a - 1;

  for(j = a; j < b; j++) {
    if(A[j] <= y) {
      i++;
      t = A[i];
      A[i] = A[j];
      A[j] = t;
    }
  }

  t = A[i + 1];
  A[i + 1] = A[b];
  A[b] = t;

  return i + 1;
}