Codecraft-17 and Codeforces Round #391 (Div. 1 + Div. 2, combined) -- B. Bash's Big Day （唯一分解定理）

大体题意：

给你n 个数，让你选出尽可能多的数来，使得gcd(x1,x2,x3,,,xn) != 1; 输出最大数量？

思路：

思路比较偏比较麻烦，参考一下把！

gcd不是1的话，说明它们有共同的质因子！

数据范围是 每个数都是10W以内！  素数也就1W个

因此我们可以给每一个数进行质因子分解。

给分解出来的质因子p   统计一下，vos[p]++，表示有一个数 是p这个素因子。

最后统计这1W个素数的最大值即可！

有两个小坑把，都是自己挖的：

这个题肯定会选一个的，因此，最小值是1，不是0，但是因为数据有1，如果不特判的话，最小值是0，因此 给答案ans初始化为1即可。

另一个这样做会超时，因为后台数据可能有 10W个超大素数（接近10W的素数），这样边枚举输入边枚举素数会超时，因此要在质因子分解中时刻判断是否是素数，是素数就不用筛了， 直接return了，这样能快了很多。

详细见代码：

#include <bits/stdc++.h>#define fi first#define se second#define ps push_back#define mr make_pairusing namespace std;const int inf = 0x3f3f3f3f;typedef long long ll;typedef unsigned long long LLU;const double eps = 1e-10;const double pi = acos(-1.0);const int maxn = 100000 + 10;int a[maxn];vector<int>pri;int vis[maxn];int vos[maxn];void init(){    int len = sqrt(100001) + 1;    for (int i = 2; i <= len; ++i) if (!vis[i])        for (int j = i*i; j <= 100000; j += i) vis[j] = 1;    for (int i = 2; i <= 100000; i++) if (!vis[i])pri.push_back(i);}void solve(int x){    for (int i = 0; i < pri.size(); ++i){        int p = pri[i];        if (!vis[x] || x == 1) {            vos[x]++;            return;        }        if (x % p == 0){            while(x % p == 0){                x/=p;            }            vos[p]++;        }    }}int main(){    init();    int n;    scanf("%d",&n);    for (int i = 0; i < n; ++i){        int x;        scanf("%d",&x);        a[i] = x;        solve(x);    }    int ans = 0;        for (int i = 0; i < pri.size(); ++i){            int p = pri[i];            ans=  max(ans,vos[p]);        }        if (!ans) ++ans;        printf("%d\n",ans);    return 0;}

B. Bash's Big Day

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

Bash has set out on a journey to become the greatest Pokemon master. To get his first Pokemon, he went to Professor Zulu's Lab. Since Bash is Professor Zulu's favourite student, Zulu allows him to take as many Pokemon from his lab as he pleases.

But Zulu warns him that a group of k > 1 Pokemon with strengths {s1, s2, s3, ..., sk} tend to fight among each other if gcd(s1, s2, s3, ..., sk) = 1 (see notes for gcd definition).

Bash, being smart, does not want his Pokemon to fight among each other. However, he also wants to maximize the number of Pokemon he takes from the lab. Can you help Bash find out the maximum number of Pokemon he can take?

Note: A Pokemon cannot fight with itself.

Input

The input consists of two lines.

The first line contains an integer n (1 ≤ n ≤ 105), the number of Pokemon in the lab.

The next line contains n space separated integers, where the i-th of them denotes si (1 ≤ si ≤ 105), the strength of the i-th Pokemon.

Output

Print single integer — the maximum number of Pokemons Bash can take.

Examples

input

32 3 4

output

2

input

52 3 4 6 7

output

3

Note

gcd (greatest common divisor) of positive integers set {a1, a2, ..., an} is the maximum positive integer that divides all the integers {a1, a2, ..., an}.

In the first sample, we can take Pokemons with strengths {2, 4} since gcd(2, 4) = 2.

In the second sample, we can take Pokemons with strengths {2, 4, 6}, and there is no larger group with gcd ≠ 1.