description
【answer】
The first number as a reference, and then [l + 1, r] be divided.
Found largest j, such that a [j] <= x, i.e., j + 1..r is greater than x
and j is this time division result
added in front of each template function
[Code]
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 1e5;
int n;
int a[N+10];
template <class Type>
int _partition(Type a[],int l,int r){
int i = l,j = r+1;
Type x = a[l];
while (i<j){
while (a[++i]<x && i<r);//a[i]最后>=x
while (a[--j]>x);
if (i>=j) break;
swap(a[i],a[j]);
}
//a[j]最后<=x,且j是最大的满足a[j]<=x的j
a[l] = a[j];
a[j] = x;
return j;
}
template <class Type>
void QuickSort(Type a[],int l,int r){
int index = _partition(a,l,r);
if (l<index) QuickSort(a,l,index-1);
if (index<r) QuickSort(a,index+1,r);
}
int main(){
//freopen("D:\\rush.txt","r",stdin);
scanf("%d",&n);
for (int i = 1;i <= n;i++){
scanf("%d",&a[i]);
}
QuickSort(a,1,n);
for (int i = 1;i <= n;i++) printf("%d ",a[i]);
return 0;
}