Less or Equal CodeForces - 977C （水题）

最新推荐文章于 2021-01-16 23:16:21 发布

Levi——

最新推荐文章于 2021-01-16 23:16:21 发布

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该博客介绍了CodeForces的一道编程题目，要求在给定长度为n的整数序列中找到一个1到10^9之间的数x，使得序列中有且仅有k个元素小于等于x。博主分享了解题思路，即先对序列排序，然后判断第m-1个元素是否等于第m个元素，以此来确定是否存在满足条件的x。当m为0时，答案为-1。

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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 $[1; 10^{9}]$ (i.e. $1 \leq x \leq 10^{9}$ ) 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 $k$ ( $1 \leq n \leq 2 \cdot 10^{5}$ , $0 \leq k \leq n$ ). The second line of the input contains $n$ integer numbers $a_{1}, a_{2}, \dots, a_{n}$ ( $1 \leq a_{i} \leq 10^{9}$ ) — the sequence itself.

Output

Print any integer number $x$ from range $[1; 10^{9}]$ 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
inputCopy
7 4
3 7 5 1 10 3 20
outputCopy
6
inputCopy
7 2
3 7 5 1 10 3 20
outputCopy
-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：给出n和m，在[1,1e9]中，找出一个数x，使得在给出的长为n的数列中小于或等与x的数只有m个。如果有，输出这个数x，否则，输出-1，答案不唯一.

思路：排序一下，然后直接看a[m-1] 和a[m] 是否相同，如果相同，那么必定不存在这样一个x，否则，a[m-1]，这个数，一定是一个满足条件的x。（注意，需要特判一下m为0的情况！！）

#include "iostream"
#include "algorithm"
using namespace std;
const int Max=2e5+10;
int a[Max];
int main()
{
    ios::sync_with_stdio(false);
    int n,m;
    cin>>n>>m;
    for(int i=0;i<n;i++) cin>>a[i];
    sort(a,a+n);
    if(m==0){
        if(a[0]<=1) cout<<"-1"<<endl;
        else cout<<"1"<<endl;
    }
    else{
        if(a[m-1]==a[m]) cout<<"-1"<<endl;
        else cout<<a[m-1]<<endl;
    }
    return 0;
}