LCPC11B - Co-Prime

容易发现答案为 $\sum \limits_{i=1}^b [\gcd(i,n)=1]-\sum \limits_{i=1}^{a-1} [\gcd(i,n)=1]$。

所以我们要解决的就是 $\sum \limits_{i=1}^m [\gcd(i,n)=1]$ 这样的问题。

我们拆柿子：

$\begin{aligned} &\; \; \; \; \sum \limits_{i=1}^m [\gcd(i,n)=1] \\&= \sum \limits_{i=1}^m \sum \limits_{d|\gcd(i,n)} \mu(d) \\ &=\sum \limits_{i=1}^m \sum \limits_{d|i \land d|n} \mu(d) \\ &= \sum \limits_{d|n} \mu(d) \sum \limits_{i=1}^m [d|i] \\ &= \sum \limits_{d|n} \mu(d) \lfloor \frac{m}{d}\rfloor\end{aligned}$

于是我们考虑枚举 $n$ 的因数，但是由于 $n \leq 10^9$，我们直接杜教筛即可。

#include <bits/stdc++.h>
using namespace std;

#define int long long

const int N = 5e5 + 5;
unordered_map<int, int> res;
int prime[N], cnt, mu[N], s_mu[N];
bool vis[N];
int t, n;

void Init()
{
	mu[1] = 1;
	for (int i = 2; i < N; i++)
	{
		if (!vis[i])
		{
			prime[++cnt] = i;
			mu[i] = -1;
		}
		for (int j = 1; j <= cnt && i * prime[j] < N; j++)
		{
			vis[i * prime[j]] = 1;
			if (i % prime[j] == 0) break;
			mu[i * prime[j]] -= mu[i];
		}
	}
	for (int i = 1; i < N; i++) s_mu[i] = s_mu[i - 1] + mu[i];
}

int get_S(int x)
{
	if (x < N) return s_mu[x];
	if (res.count(x)) return res[x];
	int ress = 1;
	for (int l = 2, r; l <= x; l = r + 1)
	{
		r = x / (x / l);
		ress -= get_S(x / l) * (r - l + 1);
	}
	return res[x] = ress;
}

inline int get_mu(int x)
{
	return get_S(x) - get_S(x - 1);
}

inline int solve(int m, int n)
{
	int ress = 0LL;
	for (int i = 1; i * i <= n; i++)
	{
		if (n % i == 0)
		{
			ress += get_mu(i) * (m / i);
			if (n / i != i)
			{
				ress += get_mu(n / i) * (m / (n / i));
			}
		}
	}
	return ress;
}

signed main()
{
	Init();
	scanf("%lld", &t);
	int c = 1;
	while (t--)
	{
		int a, b, n;
		scanf("%lld%lld%lld", &a, &b, &n);
		printf("Case #%lld: %lld\n", c++, solve(b, n) - solve(a - 1, n));
	}
	return 0;
}