CodeForces 275C k-Multiple Free Set

C. k-Multiple Free Set

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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, ..., a_n (1 ≤ a_i ≤ 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, ..., a_n}.

Sample test(s)

input

6 2
2 3 6 5 4 10

output

3

Note

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

二分查找

（寫得有點醜！）

#include<stdio.h>
#include<algorithm>
using namespace std;
long long a[100010],vis[100010];
long long ans,n,k;
void search(long long x)
{
  long long l=1,r=n;
  while(l<r)
  {
    if(a[l]==x)
    {
      if(!vis[l])
      {
        vis[l]=1;
        ans--;
        break;
      }
    }
    if(a[r]==x)
    {
      if(!vis[r])
      {
        vis[r]=1;
        ans--;
        break;
      }
    }
    long long mid=(l+r)/2;
    if(a[mid]>x)
    {
      r=mid;
    }
    else if(a[mid]<x)
    {
      l=mid;
    }
    else if(a[mid]==x)
    {
      if(!vis[mid])
      {
        vis[mid]=1;
        ans--;
        break;
      }
    }
    if(l+1==r)break;
  }
}
int main()
{
  scanf("%I64d%I64d",&n,&k);
  for(long long i=1;i<=n;i++)
  {
    scanf("%I64d",&a[i]);
  }
  if(k==1&&n>1)
  {
    printf("%I64d\n",n);
    return 0;
  }
  sort(a+1,a+n+1);
  for(long long i=1;i<=n;i++)vis[i]=0;
  ans=n;
  for(long long i=1;i<=n;i++)
  {
    if(!vis[i])search(a[i]*k);
  }
  printf("%I64d\n",ans);
  return 0;
}