从1到s选出k个数 他们的最大公约数大于1 求方案数

容斥 S(1)-S(2)+S(3) S(x)为选出k个数的公因子个数为x的数量

#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
typedef long long LL;
const int maxn = 55;
int prime[maxn], vis[maxn];
int n, m;
int get_primes(int n)
{
	int m = sqrt(n+0.5), c = 0;
	for(int i = 2; i <= n; i++)
		if(!vis[i])
		{
			prime[c++] = i;
			for(int j = i*2; j <= n; j += i)
			{
				vis[j] = 1;
			}
		}
	return c;
}
LL cm(int n, int m)
{
	LL ans = 1;
	for(int i = 1; i <= m; i++)
	{
		ans *= (LL)n--;
		ans /= (LL)i;
	}
	return ans;
}
void dfs(int i, int x, int k, int m, LL& ans, int c)
{
	if(i == c)
	{
		if(!k)
			return;
		if(k&1)
			ans += cm(m/x, n);
		else
			ans -= cm(m/x, n);
		return;
	}
	dfs(i+1, x*prime[i], k+1, m, ans, c);
	dfs(i+1, x, k, m, ans, c);
}

LL cal(int c)
{
	LL ans = 0;
	dfs(0, 1, 0, m, ans, c);
	if(ans > 10000)
		ans = 10000;
	return ans;
}
int main()
{
	int c = get_primes(50);
	//printf("%d\n", c);
	while(scanf("%d %d", &n, &m) != EOF)
	{	
		printf("%lld\n", cal(c));
	}
	return 0;
}

URAL 1091. Tmutarakan Exams 容斥