codeforces 977C Less or Equal

题目

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. $1\le x\le 10^9$) 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 $k$ ($1 \le n \le 2⋅10^5$, $0 \le k \le n$). The second line of the input contains n integer numbers $a1,a2,…,an$ ($1\le a_i \le 10^9$) — 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.

分析

【题意】
给$n$个数，输出一个$x$，满足：

• $x\in [1, 10^9]$
• 恰好有$k$个数小于等于$x$

【思路】
维护一个“比$x$小的数的集合”，不断加入更大的数并更新$x$，直到集合有$k$个数为止。一旦加入了一个数，那么和它相等的数也要一并加入，所以集合的增长是不连续的。
【注意】
需要特判$k$为0的情况，此时$x$必须比最小的数还要小并且$x\ge1$，所以最小的数必须大于1

$k$为$n$的情况，只需要选一个比最大的数还大的数就行了

代码

#include<stdio.h>
#include<algorithm>
using std::sort;
#define N_max 200005

int arr[N_max];
int main() {
    int n,k;
    scanf("%d %d", &n, &k);
    for (int i = 0; i < n;++i) 
        scanf("%d", &arr[i]);

    sort(arr, arr + n);
    //循环的边界
    arr[n] = 1000000001;
    //cur之前的数已经加入集合
    int cur = 0;
    //特判k==0
    if (k == 0) {
        if (arr[0] <= 1)  printf("-1");
        else printf("%d",arr[0]-1); return 0;
    }
    //加入第一个数以及所有与它相等的数
    while (arr[cur] == arr[cur + 1])
            cur ++;
    for ( ;;) {
        if (cur + 1 == k){printf("%d", arr[cur+1]-1);break;}
        if (cur + 1 > k) {printf("-1"); return 0;}
        //覆盖到下一个连续段的末尾
        cur++;
        while (arr[cur] == arr[cur + 1])
            cur ++;
    }
}