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

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}.

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：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;

int n, k;
long long a[100005];
bool tag[100005];

int main()
{
    while(scanf("%d %d", &n, &k) != EOF)
    {
        for(int i = 0; i < n; i++) scanf("%I64d", a + i);
        memset(tag, false, sizeof(tag));
        sort(a, a + n);
        int ans = n;
        for(int i = 0; i < n; i++)
        {
            if(!tag[i])
            {
                long long p = a[i] * k;
                if(p > a[n-1]) continue;
                int low = i + 1, high = n - 1, mid;
                while(low <= high)
                {
                    mid = (low + high) / 2;
                    if(a[mid] > p) high = mid - 1;
                    else if(a[mid] < p) low = mid + 1;
                    else
                    {
                        tag[mid] = true;
                        ans--;
                        break;
                    }
                }
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}