Codeforces #479 (Div. 3)-C. Less or Equal-

C. Less or Equal

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.

题意：
1~10的九次方 范围内的任何整数x（即1≤x≤10^9），使得给定序列的恰好k个元素小于或等于x。
让你找出是否存在x；如果不存在则输出-1；
情况是：
1.当第排序好后的第k-1个与第k个元素相等时必然不满足 否则输出a[k-1]即可；
2.当k等于0 时 给定的序列里如果a[0]等于1 必然不存在；否则 直接输出1 即可；
3.当k==n时 直接输出a[k-1]即可；
解题 排序 考虑所有出现的情况 即可

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdio>
#define ll long long
using namespace std;
const int64_t MM=1e6+5;
int  n,m;
int  a[MM];
int main()
{
    scanf("%d %d",&n,&m);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }
    sort(a,a+n);//排序
    if(m==0)
    {
        if(a[0]==1)
            printf("-1\n");
        else
            cout<<"1"<<endl;
    }
    else if(m==n)
    {
        cout<<a[m-1]<<endl;
    }
    else {
        if(a[m]==a[m-1])
            cout<<"-1"<<endl;
        else
            cout<<a[m-1]<<endl;
    }
    return 0;
}