AcWing 786. 第k个数

Posted 2022-02-09 MangataTS

题目链接

https://www.acwing.com/problem/content/788/

思路

先用一种$nlo{g}_{2}n$的排序算法排序，然后输出第k个元素即可

代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000009
#define endl "\\n"
#define PII pair<int,int>
ll ksm(ll a,ll b) 
	ll ans = 1;
	for(;b;b>>=1LL) 
		if(b & 1) ans = ans * a % mod;
		a = a * a % mod;
	
	return ans;


ll lowbit(ll x)return -x & x;

const int N = 2e6+10;
int n,a[N];

void mysort(int a[],int l,int r) 
	if(l >= r) return;
	int x = a[l + r >> 1],i = l - 1,j = r + 1;
	while(i < j) 
		while(a[++i] < x);
		while(a[--j] > x);
		if(i < j) swap(a[i],a[j]);
	
	mysort(a,l,j);
	mysort(a,j+1,r);


int main()

	int k;
	scanf("%d%d",&n,&k);
	for(int i = 0;i < n; ++i) 
		scanf("%d",&a[i]);
	
	mysort(a,0,n-1);
	printf("%d\\n",a[k-1]);
	return 0;