HDU 6231

K-th Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 599    Accepted Submission(s): 234

Problem Description

Alice are given an array A[1..N] with N numbers.

Now Alice want to build an array B by a parameter K as following rules:

Initially, the array B is empty. Consider each interval in array A. If the length of this interval is less than K, then ignore this interval. Otherwise, find the K-th largest number in this interval and add this number into array B.

In fact Alice doesn't care each element in the array B. She only wants to know the M-th largest element in the array B. Please help her to find this number.

 

Input

The first line is the number of test cases.

For each test case, the first line contains three positive numbers N(1≤N≤105),K(1≤K≤N),M. The second line contains N numbers Ai(1≤Ai≤109).

It's guaranteed that M is not greater than the length of the array B.

 

Output

For each test case, output a single line containing the M-th largest element in the array B.

 

Sample Input

2

5 3 2

2 3 1 5 4

3 3 1

5 8 2

 

Sample Output

3

2

 

Source

2017中国大学生程序设计竞赛-哈尔滨站-重现赛（感谢哈理工）

题意：

大小为n的a数组，把其中所有的长度不小于k的区间中第k大的数加入到b数组，最后求b数组中的m大的数

输入：

n,k,m

a[1~n];

代码：

//m要用long long 啊。
//二分答案x，然后尺取。找到有多少个区间存在至少k个大于等于x的数，如果这样的区间数不少于m个就说明第m大的数
//比x大，因此有单调性。尺取区间[l,r]中有k个不小于x的数那么会有n-r+1个符合的区间。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
const int MAXN=;
int n,k,a[MAXN];
ll m;
bool solve(int x)
{
    ll sum=;
    int tmp=,l=,r=;
    while(){
        while(tmp<k){
            ++r;
            if(r>n) break;
            if(a[r]>=x) tmp++;
        }
        if(r>n) break;
        sum+=(n-r+);
        if(a[l]>=x) tmp--;
        l++;
    }
    if(sum>=m) return ;
    return ;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d%lld",&n,&k,&m);
        for(int i=;i<=n;i++) scanf("%d",&a[i]);
        int l=,r=1e9,ans=;
        while(l<=r){
            int mid=(l+r)>>;
            if(solve(mid)) { ans=mid;l=mid+; }
            else r=mid-;
        }
        printf("%d\n",ans);
    }
    return ;
}