本文主要是介绍CodeForces - 977 C. Less or Equal，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
You are given a sequence of integers of length nn and integer number kk. You should print any integer number xx in the range of [1;109][1;109] (i.e. 1≤x≤1091≤x≤109) such that exactly kk elements of given sequence are less than or equal to xx.
Note that the sequence can contain equal elements.
If there is no such xx, print "-1" (without quotes).
Input
The first line of the input contains integer numbers nn and kk (1≤n≤2⋅1051≤n≤2⋅105, 0≤k≤n0≤k≤n). The second line of the input contains nn integer numbers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) — the sequence itself.
Output
Print any integer number xx from range [1;109][1;109] such that exactly kk elements of given sequence is less or equal to xx.
If there is no such xx, 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 55 is also a valid answer because the elements with indices [1,3,4,6][1,3,4,6] is less than or equal to 55 and obviously less than or equal to 66.
In the second example you cannot choose any number that only 22 elements of the given sequence will be less than or equal to this number because 33 elements of the given sequence will be also less than or equal to this number.
此题不难，主要是k==0时的讨论想明白就OK了，卡了好久，伤心
#include<bits/stdc++.h>
using namespace std;
long long a[2000008];
int main()
{int n,k,ans=0,temp;scanf("%d%d",&n,&k);for(int i=1;i<=n;i++)scanf("%lld",&a[i]);sort(a+1,a+1+n);if(k==0){if(a[1]>1)printf("1\n");elseprintf("-1\n");return 0;}if((a[k]!=a[k+1]&&k<n)||n==k)printf("%lld\n",a[k]);else printf("-1\n");return 0;
}
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