Codeforces Round #168 (Div. 2) C. kMultiple Free Set（二分查找）

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

Note

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

数据需要用64位long long去存储。

先对原始数据排序，遍历数组，若某元素的k倍数存在，则要判断是保留该元素抑或他的k倍数，事实上，总是保留该元素，然后标记它的k倍数，之后不再检测。

在判断k倍数是否存在时，使用顺序查找会超时，使用二分查找即可。

AC Code：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 
 6 using namespace std;
 7 
 8 int n, k;
 9 long long a[100005];
10 bool tag[100005];
11 
12 int main()
13 {
14     while(scanf("%d %d", &n, &k) != EOF)
15     {
16         for(int i = 0; i < n; i++) scanf("%I64d", a + i);
17         memset(tag, false, sizeof(tag));
18         sort(a, a + n);
19         int ans = n;
20         for(int i = 0; i < n; i++)
21         {
22             if(!tag[i])
23             {
24                 long long p = a[i] * k;
25                 if(p > a[n-1]) continue;
26                 int low = i + 1, high = n - 1, mid;
27                 while(low <= high)
28                 {
29                     mid = (low + high) / 2;
30                     if(a[mid] > p) high = mid - 1;
31                     else if(a[mid] < p) low = mid + 1;
32                     else
33                     {
34                         tag[mid] = true;
35                         ans--;
36                         break;
37                     }
38                 }
39             }
40         }
41         printf("%d\n", ans);
42     }
43     return 0;
44 }