Code forces 275C-----思维---二分

k-Multiple Free Set

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 274A

Description

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 Input

Input

6 22 3 6 5 4 10

Output

3

这个题感觉还是公式，找规律应该可以找到，应该会非常非常麻烦，直接用规律做了。就是从小到大遍历，某个数和它的k倍不能同时出现，总是把它的k倍踢出去才是最优解，数学真奇妙---

因为数据多而且大，所以用二分了，否则超时

#include<cstdio>#include<cmath>#include<algorithm>#include<cstring>using namespace std;#define maxn 500005#define eps 2e12__int64 a[maxn];bool b[maxn];int main(){int n, k;__int64 s, e, m, cnt;scanf("%d%d", &n, &k);for(int i = 0; i < n; i++){scanf("%I64d", &a[i]);}cnt = n;memset(b, true, sizeof(b));sort(a, a + n);for(int i = 0; i < n; i++){if(b[i]){s = i+1;e = n-1;if(a[i]*k > a[e]){break;}while(e >= s){m = (e+s) >> 1;if(a[m] > k*a[i]){e = m-1;}else if(a[m] < k*a[i]){s = m+1;}else{b[m] = false;cnt--;break;}}}}printf("%I64d\n", cnt);return 0;}