K-th Number

标签：fir   --   include   line   序列   ISE   syn   nta   ==   

题意:

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.

输入：

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.

输出:

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

 样例输入：

2

5 3 2

2 3 1 5 4

3 3 1

5 8 2

输出：

3

2

题目大意：给定一个序列A，将序列的每个区间（区间长度>=K）中第K大的数加入到B数组中，求B数组中第M大的数。

解题思路：二分第M大的值？ 当值为X时，统计出第K大时>=X的区间个数。 如果区间个数大于M，那么答案肯定比X小。

#include <bits/stdc++.h>
#define LL long long
using namespace std;
const int N=1e5+5;
int a[N],n,k;
LL m;
int check(int x)
{
    LL ans=0;
    int j=1,num=0;
    for(int i=1;i<=n;i++){
        if(a[i]>=x) num++;
        if(num==k){
            ans+=n-i+1;
            while(a[j]<x){
                ans+=n-i+1;
                j++;
            }
            j++;
            num--;
        }
    }
    if(ans>=m) return true;
    return false;
}
int main()
{
    int T;
    ios::sync_with_stdio(false);
    cin>>T;
    while(T--){
        cin>>n>>k>>m;
        for(int i=1;i<=n;i++) cin>>a[i];
        int L=1,R=1000000000,ans=0;
        while(L<=R){
            int mid=(L+R)>>1;
            if(check(mid)) ans=mid,L=mid+1;
            else R=mid-1;
        }
        cout<<ans<<endl;
    }
}

K-th Number

标签：fir   --   include   line   序列   ISE   syn   nta   ==   

原文地址：https://www.cnblogs.com/ww123/p/11632025.html