HDU6231 K-th Number（二分答案+双指针）

题目：传送门

思路：二分答案x，用双指针判断 区间个数，区间满足至少有k个数>=x ，再根据个数和m的大小关系继续二分。易得，最后二分得到的 ans 一定是数组中的某一个数，否则就会继续二分。

#include<bits/stdc++.h>
/*
#include<cstdio>
#include<cmath>
#include<cstring>
#include<vector>
#include<cctype>
#include<queue>
#include<algorithm>
#include<map>
#include<set>
*/
#pragma GCC optimize(2)
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
typedef pair<double,double> pdd;
const int N=1e6+5;
const int M=1e4+5;
const int inf=0x3f3f3f3f;
const LL mod=1e9+7;
const double eps=1e-9;
const long double pi=acos(-1.0L);
#define ls (i<<1)
#define rs (i<<1|1)
#define fi first
#define se second
#define pb push_back
#define mk make_pair
#define mem(a,b) memset(a,b,sizeof(a))
LL read()
{
    LL x=0,t=1;
    char ch;
    while(!isdigit(ch=getchar())) if(ch=='-') t=-1;
    while(isdigit(ch)){ x=10*x+ch-'0'; ch=getchar(); }
    return x*t;
}
int n,k,a[N];
LL m;
int judge(int x)
{
    LL res=0,sum=0;
    for(int l=1,r=1;l<=n;l++)
    {
        while(sum<k&&r<=n) sum+=a[r++]>=x;
        if(sum==k) res+=n-r+2;
        sum-=a[l]>=x;
    }
    return res>=m;
}
int main()
{
    int T=read();
    while(T--)
    {
        n=read(),k=read(),m=read();
        for(int i=1;i<=n;i++) a[i]=read();
        int l=1,r=1e9;
        while(l<=r)
        {
            int mid=l+r>>1;
            if(judge(mid)) l=mid+1;
            else r=mid-1;
        }
        printf("%d
",r);
    }

    return 0;
}