【Codeforces 757 Bash's Big Day】+ 手速

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

3
2 3 4

Output

2

Input

5
2 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.

统计每个因子最多能整除的数的个数～最大 1e5 =-=～O（n × ln n）

AC代码：

#include<bits/stdc++.h>using namespace std;int ma[100010];int main(){    int N,a;    scanf("%d",&N);    for(int i = 1; i <= N; i++){        scanf("%d",&a);        for(int j = 1 ; j * j <= a ; j++)            if(a % j == 0){               ma[j]++;            if(j * j != a)                ma[a / j]++;            }    }    int cut = 1;    for(int i = 2 ; i < 100001; i++)        cut = max(cut,ma[i]);    printf("%d\n",cut);    return 0;}