Codeforces977C.Less or Equal 1200(k=0坑)

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探讨了如何在给定的整数序列中找到一个整数x，使得恰好k个元素小于或等于x，范围限制在[1,10^9]内。如果不存在这样的x，则返回-1。通过排序和比较解决了这一问题。

You are given a sequence of integers of length n and integer number k. You should print any integer number x in the range of [1;109] (i.e. 1≤x≤109) such that exactly k elements of given sequence are less than or equal to x.

Note that the sequence can contain equal elements.

If there is no such x, print “-1” (without quotes).

Input
The first line of the input contains integer numbers n and k (1≤n≤2⋅10⁵, 0≤k≤n). The second line of the input contains n integer numbers a1,a2,…,an (1≤ai≤10⁹) — the sequence itself.

Output
Print any integer number x from range [1;10⁹] such that exactly k elements of given sequence is less or equal to x.

If there is no such x, print “-1” (without quotes).

Examples
input
7 4
3 7 5 1 10 3 20
output
6
input
7 2
3 7 5 1 10 3 20
output
-1
Note
In the first example 5 is also a valid answer because the elements with indices [1,3,4,6] is less than or equal to 5 and obviously less than or equal to 6.

In the second example you cannot choose any number that only 2 elements of the given sequence will be less than or equal to this number because 3 elements of the given sequence will be also less than or equal to this number.

注意: k的范围 k可以为0 序列中0个数<=x(1升)
k=0 最小数a[1]=1 立马1个数<=1 与k=0矛盾

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<cmath>
#include<cctype>
using namespace std;

#define PI acos(-1.0)
#define mp make_pair
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

const int INF32M=0x3f3f3f3f;
const ll INF64M=0x3f3f3f3f3f3f3f3f;
const int maxn=2e5+5;
int a[maxn];
int main()
{
	ios::sync_with_stdio(false);
 	int n,k;//n<=2e5 k最多拿序列中2e5个元素 输出x在1-1e9不会爆int 
 	cin>>n>>k;
 	for(int i=1;i<=n;i++)
 		cin>>a[i];
 	sort(a+1,a+n+1);
 	if(k==0)	//n=3 k=0 0个数小于等于x(x 1-1e9) 
 	{
 		//a[i] 1-1e9 min=1 
 		if(a[1]==1)	//最坏情况x取最小=1 0个数<=1 值为1 立刻存在 1个数<=1 
 			cout<<-1<<endl;
 			//保证最小a[1]的比1大 
 		else	//n=3 k=0 2 3 4	0个数小于等于x 没有比x更小的数 2起步 
 			cout<<1<<endl;//cout<<a[1]-1<<endl; x
 			
	}
	else//n=7 k=2    3 7 5 1 10 3 20 存在k=2个数<=x 
	//1 3 3 5 7 10 20 要求两个数<=3(4) 实际上3个数<=3(4)
	{
		if(a[k]==a[k+1])	//多一个都不行
			cout<<-1<<endl;	//与要求k不符
		else
			cout<<a[k]<<endl;//n=7 k=4	   3 7 5 1 10 3 20  
 	//1 3 3 5 7 10 20
			
	}
    return 0;
}