#include
    
     using namespace std;int a[100];void swap(int &a, int &b)	{		int t = a;		a = b;		b = t;	}void merge(int a[],int s,int m, int e, int tmp[])//归并 把每一小部分归为有序 	{		int pt = 0;		int p1 = s,p2 = m+1;		while(p1 <= m && p2 <= e){			if(a[p1] < a[p2]){				tmp[pt++] = a[p1++];			}			else{				tmp[pt++] = a[p2++];			}		}		while(p1 <= m)//把剩余可能部分的加入归并结果 			tmp[pt++] = a[p1++];		while(p2 <= e)			tmp[pt++] = a[p2++];		for(int i = 0; i < e-s+1; ++i){//把此部分归并（排序）结果重新赋值到a中 			a[s+i] = tmp[i];		}		}void mergeSort(int a[],int s, int e,int tmp[])	{		if(s < e){			int m = s + (e-s)/2;//记录中点 			mergeSort(a,s,m,tmp);			mergeSort(a,m+1,e,tmp);			merge(a,s,m,e,tmp);		}	}	void arrangeRight(int *a, int s, int e, int k)//利用线性时间找出前k大的数放在右边{    if(s >= e)        return;    if(k == e-s+1)        return;    int i = s;    int j = e;    int key = a[s];    while(i != j)    {         while(i < j && a[j] >= key)             --j;         swap(a[i],a[j]);         while(i < j && a[i] <= key)             ++i;         swap(a[i],a[j]);    }    //如果右边的数刚好为e-i+1为k个    if(k == e - i + 1)        return;    //右边的数大于k个    else if(k < e - i + 1)    {        arrangeRight(a,i+1,e,k);    }    //右边的数小于k个，需要左边取出k - (e-i+1)个    else    {         arrangeRight(a,s,i-1,k-e+i-1);    }}int main()	{		int a[] = {4,3,1,6,9,11,5,8,10,2};		int tmp[11];		int n = sizeof(a)/sizeof(int);		int m;		scanf("%d",&m);		arrangeRight(a,0,n-1,m);		mergeSort(a,n-m,n-1,tmp);		for(int i = 0; i < n; ++i){			printf("%4d",a[i]);		}				return 0;	}