K-th Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 635 Accepted Submission(s): 247

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

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）

最近感觉遇到瓶颈了

可能还是太贪玩了吧=0=

给你三个数字 n, k, m (我的变量命名为 n m w= =)

题意看起来挺乱

将所有长度大于 k 的区间的第 k 大的数提取出来

丢到一个新的数组里B

然后输出这个数组B的第 m 大

因为 n 为 1e5

所以数组 B 中最多会有 1e10个元素

所以肯定不可能直接模拟

那么我们考虑二分答案

对于二分出来的 mid

我们统计出区间的第 k 大大于 mid 的区间个数 num

如果 num 大于 m

那就说明 mid 太小了

反之则说明 mid 太大了

接着就剩下怎么 O(n)的找出 num

考虑尺取

cnt 表示 [l, r]区间内大于等于 mid 的数字个数

当 cnt < k

移动 r

当cnt == k的时候

说明

[l, r] ~ [l, n - 1] (因为我的下标从 0 开始)的区间第 k 大都是大于 mid 的

所以 num += n - r

然后移动 l

重新回想起来了二分的精髓 —— 根本不需要在意是如何算出来的只需要无限的缩小答案的选择区间就好了

#include <bits/stdc++.h>
#define LL long long
using namespace std;

const int N = 100010;

LL n, m, w;
int ma[N];
int mb[N];

bool check(int x)
{
    int l;
    int r;
    int cnt;
    LL sum;

    cnt = 0;
    sum = 0;
    l = 0;
    r = -1;
    while(r < n){
        if(cnt < m){
            r ++;
            if(ma[r] >= x){
                cnt ++;
            }
        }
        else{
            sum += (n - r);
            if(ma[l] >= x){
                cnt --;
            }
            l ++;
        }
    }
    if(sum >= w){
        return true;
    }
    else{
        return false;
    }
}

int main()
{
    int ncase;
    int l, r;
    int ans;

    scanf("%d", &ncase);
    while(ncase --){
        scanf("%d%d%d", &n, &m, &w);
        for(int i = 0; i < n; i ++){
            scanf("%d", &ma[i]);
            mb[i] = ma[i];
        }
        sort(mb, mb + n);
        l = 0;
        r = n - 1;
        while(l <= r){
            int mid = l + ((r - l) >> 1);
            int t;

            t = mb[mid];
            if(check(t)){
                ans = t;
                l = mid + 1;
            }
            else{
                r = mid - 1;
            }
        }
        printf("%d\n", ans);
    }

    return 0;
}

[HDU] 6231 - 哈尔滨 - 感谢哈理工 - K-th Number - 二分 + 尺取

K-th Number

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