[二分]codeforces 274A k-Multiple Free Set

A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.

You're given a set of n distinct positive integers. Your task is to find the size of it's largest k-multiple free subset.

Input

The first line of the input contains two integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ 109). The next line contains a list of n distinct positive integers a1, a2, ..., an (1 ≤ ai ≤ 109).

All the numbers in the lines are separated by single spaces.

Output

On the only line of the output print the size of the largest k-multiple free subset of {a1, a2, ..., an}.

Examples

input

Copy

6 2
2 3 6 5 4 10

output

Copy

Note

In the sample input one of the possible maximum 2-multiple free subsets is {4, 5, 6}.

题意:给你n个数，找出在这n个数中最多有多少个x使得所选出的数中比x大的数不是x的k倍

注意：先选择所有的数，若x的k倍在原序列存在，则看x的k倍的k倍是否存在，若存在则删去x的k倍，因为这样就可以选x和x的k倍的k倍，若x的k倍的k倍不存在，则删x的k倍

排序后二分查找，k和a[i]最大是1e9，所以要注意k*k*a[i]是否小于1e18，为防止溢出可用if(k*a[i]<=(1e18)/k)

自己的测试数据：

2 1000000000
1000000000 1000000000

4 2
2 3 4 8

5 1
1 2 3 4 5

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int amn=1e5+5;
 5 const ll inf=1e18;
 6 
 7 int n,k,mid;
 8 ll a[amn];
 9 
10 ll fd(ll b){
11     int l=1,r=n;
12     mid=(l+r)/2;
13     while(l<r){
14         if(a[mid]<b)l=mid+1;
15         else if(a[mid]>b) r=mid-1;
16         else break;
17         mid=(l+r)/2;
18     }
19     return a[mid];
20 }
21 
22 int main(){
23     ios::sync_with_stdio(0);
24     cin>>n>>k;
25     for(int i=1;i<=n;i++){
26         cin>>a[i];
27     }
28     sort(a+1,a+1+n);
29     int ans=n,pos,jg,jg1;
30     if(k!=1)
31     for(int i=1;i<n;i++){
32         if(!a[i])continue;
33         jg=fd(k*a[i]);
34         if(k*a[i]==jg){
35             pos=mid;
36             if(k*a[i]<=inf/k){
37                 jg1=fd(k*k*a[i]);
38                 if(k*k*a[i]==jg1){
39                 a[pos]=0;
40                 }
41                 else a[i]=0;
42             }
43             else a[i]=0;
44             ans--;
45         }
46     }
47     printf("%d
",ans);
48 }