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 i,m,n;
    int A[N];

    scanf("%d",&n);

    for(i=0;i<n;i++)
      scanf("%d",&A[i]);
    
    m=partition(A,0,n-1);

    for(i=0;i<m;i++)
      printf("%d ",A[i]);

    printf("[%d] ",A[m]);

    for(i=m+1;i<n-1;i++)
      printf("%d ",A[i]);

    printf("%d\n",A[n-1]);

    return 0;

}

int partition(int *A,int p,int r){

    int x,i,j;
    int tmp1,tmp2;

    x=A[r];

    i=p-1;

    for(j=p;j<r;j++){
        if(A[j]<=x){
            i++;
            tmp1=A[i];
            A[i]=A[j];
            A[j]=tmp1;
        }
    }

    tmp2=A[i+1];
    A[i+1]=A[r];
    A[r]=tmp2;
    
    return i+1;

}