【暴力】欠扁的 CD

$Link$

$ssl1495$

$Description$

$Sample$ $Input$

3 1
1	2	3

$Sample$ $Output$

3

$Hint$

$Solution$

枚举$gcd$，然后再枚举$gcd$的倍数在原序列中出现的次数，若大于等于k就直接输出即可

$Code$

#include
#include
#include
#define ll long long

using namespace std;

ll s[500005], a[500005];
ll n, k, maxn;

ll check(int x)
{
	ll sum = 0;
	for (int i = 1; i * x <= maxn; ++i)
		sum += s[i * x];
	if (sum >= k) return 1;
	return 0;
}

int main()
{
	scanf("%lld%lld", &n, &k);
	for (int i = 1; i <= n; ++i){
		scanf("%lld", &a[i]);
		maxn = max(maxn, a[i]);
		s[a[i]]++;//记录每个数出现次数
	} 
	sort(a + 1, a + n + 1);
	for (ll i = a[n]; i >= 1; --i)
	{
		if (check(i)) {
			printf("%lld", i * k);//有k个人，每个人贡献i
			return 0;
		} 
	}
	return 0;
}