CF364D Ghd

题目链接

采用随机的思想，钦定一个一定对答案有贡献的数。把这个数的所有因子拿出来求个贡献就好。（随机 (10) 次错误的概率就是 (frac 1 {2^{10}})）。

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<random>
#include<ctime>

using namespace std;

typedef long long LL;
const int N = 1000009;
int n;
LL a[N], Ans, d[N], f[N];
mt19937 rnd(time(0));

void init()
{
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		scanf("%lld", &a[i]);
}

void work()
{
	for (int qwq = 1; qwq <= 12; qwq++)
	{
		LL k = rnd() % n + 1, m = a[k], cnt=0;
		for (LL i = 1; i * i <= m; i++)
			if(m % i == 0)
			{
				d[++cnt] = i;
				if (i * i != m)
					d[++cnt] = m / i;
			}
		sort(d + 1, d + 1 + cnt);
		memset(f, 0, sizeof(f));
		for (int i = 1; i <= n; i++)
			f[lower_bound(d + 1, d + 1 + cnt, __gcd(a[i], m)) - d]++;
		for (int i = cnt; i; i--)
		{
			int tot = 0;
			for (int j = i; j <= cnt; j++)
				if(d[j] % d[i] ==0)
					tot += f[j];
			if(tot >= (n - 1) / 2 + 1)
			{
				Ans = max(Ans, d[i]);
				break;
			}
		}
	}
	printf("%lld
", Ans);
}

int main()
{
	init();
	work();
	return 0;
}