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;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.

分析

【题意】
给 n 个数，输出一个 x ，满足：

• x∈[1,109]
• 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≥1 ，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
#include
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 ++;
    }
}