codeforces 757 B. Bash's Big Day

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 ifgcd(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.

题意:给出一个序列,选出最多的数使得这些数的gcd不为1

在哪见过相似的题，但是当时太弱被吓跑了...

尴尬只好yy了

不知道是不是正解(感觉挺乱搞的....) ：

1.线性筛素数，并记录最小质因子(以方便在log的复杂度内分解质因数)。

2.开一个set数组S[]，S[i]表示能被编号为i的质数整除的数的集合(set里存该数下标防止被去重，或者开multiset就可以任性的随便存了)

3.从1~n枚举每个数a[i]，对a[i]分解质因数，当分解到质因数p时进行如下操作：

壹：ans=max(ans，S[p(p在质数表的编号)].size() )   [表示选的第一个数是a[i],且最终选出的集合的p|gcd的集合大小

贰：S[p].erase(i)，根据 壹 中意义，a[i]不对i后的答案作贡献，从其质因子集合中删去

代码：

#include<iostream>#include<algorithm>#include<cstdlib>#include<cstring>#include<climits>#include<cstdio>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#define N 200020using namespace std;inline int read(){int x=0,f=1;char ch='#';while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;}int prime[N],cnt;bool is[N];int dy[N];void init(){for(int i=2;i<=100000;i++){if(!is[i]){prime[++cnt]=i;dy[i]=cnt;}for(int j=1;j<=cnt;j++){long long tmp=prime[j]*i;if(tmp>100000)break;is[tmp]=1;dy[tmp]=j;if(i%prime[j]==0)break;}}}int n,a[N];set<int>S[N];set<int>T[N];set<int>:: iterator it;void diver(int x,int ps){int t=x;while(x!=1){T[dy[x]].insert(ps);S[ps].insert(dy[x]);x/=prime[dy[x]];}}int ans=1;int main(){init();n=read();for(int i=1;i<=n;i++)a[i]=read();for(int i=1;i<=n;i++)diver(a[i],i);for(int i=1;i<=n;i++)for(it=S[i].begin();it!=S[i].end();it++){int x=(*it);int num=T[x].size();ans=max(num,ans);T[x].erase(i);}printf("%d\n",ans);}