Codeforces B. Micro-World

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我是靠谱客的博主爱笑洋葱，最近开发中收集的这篇文章主要介绍Codeforces B. Micro-World，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

You have a Petri dish with bacteria and you are preparing to dive into the harsh micro-world. But, unfortunately, you don't have any microscope nearby, so you can't watch them.

You know that you have $n$ bacteria in the Petri dish and size of the $i$ -th bacteria is $a_{i}$ . Also you know intergalactic positive integer constant $K$ .

The $i$ -th bacteria can swallow the $j$ -th bacteria if and only if $a_{i} > a_{j}$ and $a_{i} \leq a_{j} + K$ . The $j$ -th bacteria disappear, but the $i$ -th bacteria doesn't change its size. The bacteria can perform multiple swallows. On each swallow operation any bacteria $i$ can swallow any bacteria $j$ if $a_{i} > a_{j}$ and $a_{i} \leq a_{j} + K$ . The swallow operations go one after another.

For example, the sequence of bacteria sizes $a = [101, 53, 42, 102, 101, 55, 54]$ and $K = 1$ . The one of possible sequences of swallows is: $[101, 53, 42, 102, 101 - -- -, 55, 54]$ $\to$ $[101, 53 - - -, 42, 102, 55, 54]$ $\to$ $[101 - -- -, 42, 102, 55, 54]$ $\to$ $[42, 102, 55, 54 - - -]$ $\to$ $[42, 102, 55]$ . In total there are $3$ bacteria remained in the Petri dish.

Since you don't have a microscope, you can only guess, what the minimal possible number of bacteria can remain in your Petri dish when you finally will find any microscope.

Input

The first line contains two space separated positive integers $n$ and $K$ ( $1 \leq n \leq 2 \cdot 10^{5}$ , $1 \leq K \leq 10^{6}$ ) — number of bacteria and intergalactic constant $K$ .

The second line contains $n$ space separated integers $a_{1}, a_{2}, \dots, a_{n}$ ( $1 \leq a_{i} \leq 10^{6}$ ) — sizes of bacteria you have.

Output

Print the only integer — minimal possible number of bacteria can remain.

   Examples 
 
     input 
    
      Copy 
    
7 1
101 53 42 102 101 55 54

     output 
    
      Copy 
    
3

     input 
    
      Copy 
    
6 5
20 15 10 15 20 25

     output 
    
      Copy 
    
1

     input 
    
      Copy 
    
7 1000000
1 1 1 1 1 1 1

     output 
    
      Copy 
    
7

Note

The first example is clarified in the problem statement.

In the second example an optimal possible sequence of swallows is: $[20, 15, 10, 15, 20 - - -, 25]$ $\to$ $[20, 15, 10, 15 - - -, 25]$ $\to$ $[20, 15, 10 - - -, 25]$ $\to$ $[20, 15 - - -, 25]$ $\to$ $[20 - - -, 25]$ $\to$ $[25]$ .

In the third example no bacteria can swallow any other bacteria.

题解：整体扫一遍，用二分查找是否有符合条件的。刚开始开1e6的数组打表，结果超时了，额....。

笔记：

#include<algorithm> 中
lower_bound()为大于等于n的第一个数的位置
upper_bound()为大于n的第一个数的位置

参考博客

#include<functional> 中
greater<int>()函数即可对逆序求最近位置

注意：还有一个作用，可以在优先队列里使用，使其正序。

（其实，用cmp也可以）

在逆序排序的时候可以直接用。

#include<bits/stdc++.h>
using namespace std;
int a[1000000+100];
int main()
{
	int n,m,sum=0;
	cin>>n>>m;
	for(int i=0;i<n;i++)
	{
		cin>>a[i];
	}
	sort(a,a+n,greater<int>());
	for(int i=0;i<n;i++)
	{
		int cc=a[i]+m;
		int aum=lower_bound(a,a+n,cc,greater<int>())-a;
		if(a[aum]>a[i])
		{
			sum++;
		} 
	}
	cout<<n-sum<<endl;
	return 0;
}