CF - 757B. Bash's Big Day - 贪心+暴力

1.题目描述：

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.

2.题意概述：

给你一组数，要你求它的子集中最大公约数不为1的最大数量（注意当全为1时数量就是它本身，这个地方wa了一次真心应该好好读题）

3.解题思路：

暴力打表也要有技巧，开记忆数组num枚举它的质因子，最后统计一下就行

核心代码：

vis[j]++;

vis[num / j]++;

4.AC代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define maxn 100100
using namespace std;
int vis[maxn];
int main()
{
int n;
while (scanf("%d", &n) != EOF)
{
memset(vis, 0, sizeof(vis));
for (int i = 0; i < n; i++)
{
int num;
scanf("%d", &num);
for (int i = 1; i*i <= num; i++)
{
if (num % i == 0)
{
vis[i]++;
if (num / i != i)
vis[num / i]++;
}
}
}
int ans = 1;
for (int i = 2; i < maxn; i++)
ans = max(ans, vis[i]);
printf("%d\n", ans);
}
return 0;
}