codeforce B. Bash Big Day

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.

 

思路：把一個個數拿去因式分解，因子最多的那個就是答案。因式分解最快的方法就是把所有的素數先篩選出來，這個可以用篩法實現。

AC代碼：

 

#include<stdio.h>
#include<string.h>
using namespace std;
typedef long long int ll;
int a[100010], n, prime[100010], f[100010], Cnt[100010];
int flage[100010];
int cnt;
void Break(int r) {
	if (flage[r]) {
		Cnt[r]++;
		return;
	}
	for (int i = 0; i < cnt; i++) {
		if (r == 1)return;
		if (r%prime[i] == 0) {
			Cnt[prime[i]]++;
		}
		while (r%prime[i] == 0) {
			r = r / prime[i];
		}
	}
}
int main() {
	int i, j, k, Max=-1;
	scanf("%d", &n);
	for (i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
		if (a[i] > Max) {
			Max = a[i];
		}
	}
	cnt = 0;
	for (i = 2; i <= 100000; i++) {
		if (!f[i]) {
			f[i] = 1;
			flage[i] = 1;
			prime[cnt++] = i;
			for (j = i * 2; j <= 100000; j = j + i) {
				f[j] = 1;
			}
		}
	}
	for (i = 1; i <= n; i++) {
		Break(a[i]);
	}
	int ans = 1;
	for (i = 2; i <= Max; i++) {
		if (Cnt[i] > ans) {
			ans = Cnt[i];
		}
	}
	printf("%d", ans);
	return 0;
}