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);

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

int partition(int *a,int p,int r){
  int x,i,j,t;
  x=a[r];
  i=p-1;
  //printf("x=%d,i=%d,p=%d,r=%d\n",x,i,p,r);
  for(j=p; j<r; j++){
    if(a[j]<=x){
      i++;
      t=a[i];a[i]=a[j];a[j]=t;
    }
  }
  t=a[i+1];a[i+1]=a[r];a[r]=t;
  return i+1;
}