【CodeForces 339B】Micro-World

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题目描述

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: $[101, 53, 42, 102, \underline{101}, 55, 54]$ $\to$ $[101, \underline{53}, 42, 102, 55, 54]$ $\to$ $[\underline{101}, 42, 102, 55, 54]$ $\to$ $[42, 102, 55, \underline{54}]$ $\to$ $[42, 102, 55]$ . 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.

题目翻译（摘自luogu）

给出一段长度为n的初始数列，当a[i]>a[j]并且a[i]<=a[j]+k时，可以将a[j]删除。求最终数列可能的最小长度

输入：

第一行输入n,k

第二行输入n个数表示a[i]

输出：

最终最小的可能长度

感谢@zhaotiensn 提供翻译

解题思路

先排序，然后扫一遍。

水题。

代码

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cstring>
 5 using namespace std;
 6 int n,k;
 7 int a[500000];
 8 inline void read(register int &x){
 9     x=0; register char ch=getchar();
10     while(ch<'0'||ch>'9')ch=getchar();
11     while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
12 }
13 int main(){
14     read(n),read(k);
15     for(register int i=1;i<=n;i++){
16         read(a[i]);
17     }
18     sort(a+1,a+n+1);
19     int ans=n;
20     for(register int i=1;i<n;i++){
21         int num=a[i],pos=i;
22         while(a[i+1]==a[i])i++;
23         int ne=a[i+1];
24         //cout<<num<<' '<<ne<<endl;
25         if(ne>num&&ne<=num+k)ans-=(i-pos+1);
26     }
27     cout<<ans<<endl;
28 }

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 $K$ .

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.