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}.

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}.

解题思路：思路很简单，找出所有分组，且每个分组内的数从小到大排序满足前一个数的k倍为后一个数，然后根据这些分组的大小求解最终的结果，分奇数情况和偶数情况两种方法进行讨论。
代码实现由细节的方面，对k=1时，处理需要特别注意容易造成死循环。

#include <ctime>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <string>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;
const int maxn = 100010;
typedef long long ll;
map<ll, int> myMap;
ll arr[maxn];
bool mark[maxn];

int main() {

    //freopen("aa.in", "r", stdin);

    int n, k, ans = 0;
    scanf("%d %d", &n, &k);
    memset(mark, false, sizeof(mark));
    for(int i = 1; i <= n; ++i) {
        scanf("%I64d", &arr[i]);
    }
    sort(arr + 1, arr + n + 1);
    for(int i = 1; i <= n; ++i) {
        myMap[arr[i]] = i;
    }
    for(int i = 1; i <= n; ++i) {
        if(!mark[i]) {
            ll x = arr[i];
            int ts = 0;
            // 此处需要添加 !mark[myMap[x]] 这样一个判断条件
            // 从而避免k=1时造成的死循环
            while(myMap.find(x) != myMap.end() && !mark[myMap[x]]) {
                ts++;
                mark[myMap[x]] = true;
                x = x * k;
            }
            ans += (ts/2) + (ts%2);
        }
    }
    printf("%d\n", ans);
    return  0;
}