HDU - 6231 K-th Number （2017CCPC哈尔滨站 二分+尺取法）

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

题意：给你个数组A，将每个区间的第k大的数字加入数组B，不存在则跳过，求B中第m大的数

题解：二分答案，尺取法计当大于mid的第k大个数的个数，如果大于m，则说明答案在mid与right之间，否则在left与mid之间。（二分左闭右开）

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
#define maxn 100010
#define CLR(a,b) memset(a,b,sizeof(a))
long long n,k,m;
int a[maxn];

bool check(int mid)
{
    int cnt = ;
    int l = ;
    int r = ;
    long long sum = ;
    while(r<=n){
        if(cnt < k){
            r++;
            if(a[r] >= mid) cnt++;
        }
        if(cnt >= k){
            sum+=(n-r+);
            if(a[l] >= mid) cnt--;
            l++;
        }
    }
    return (sum>=m);
}

int main()
{
    int t;
    cin>>t;
    while(t--){
        cin>>n>>k>>m;
        int l = ;
        int r = ;
        for(int i=;i<=n;i++){
            scanf("%d",&a[i]);
            r = max(a[i]+,r);
        }
        while(l+<r){
            int mid = (r+l)/;
            if(check(mid))
                l = mid;
            else
                r = mid;
        }
        cout<<l<<endl;
    }
    return ;
}

/*

1
8 3 4
2 3 1 5 3 4 4 5

*/