HDU 6390 GuGuFishtion

題意：

計算：

$$\sum\limits_{a = 1}^{m}\sum\limits_{b = 1}^{n} \frac{\varphi(ab)}{\varphi(a)\varphi(b)} (\bmod p)$$

思路：

考慮算術基本定理和$\varphi(x)$函數積性將式子化簡：

令$a = p_1^{t_1}p_2^{t_2} \cdots p_n^{t_n}$，$b = p_1^{q_1}p_2^{q_2} \cdots p_n^{q_n}$。

那麼原式有：

$$ \begin{eqnarray*} \frac{\varphi(ab)}{\varphi(a)\varphi(b)} (\bmod p) = \frac{\varphi(p_1^{t_1 + q_1} \cdots p_n^{t_n + q_n})}{\varphi(p_1^{t_1} \cdots \varphi(p_n^{t_n})) \cdot \varphi(p_1^{q_1} \cdots p_n^{q_n})} (\bmod p) \end{eqnarray*} $$

咱們單獨考慮一下$p_1$，那麼有：

$$ \begin{eqnarray*} \frac{\varphi(p_1^{t_1 + q_1})}{\varphi(p_1^{t_1}) \cdot \varphi(p_1^{q_1})} = \frac{p_1^{t_1 + q_1} \cdot (1 - \frac{1}{p_1})} {p_1^{t_1} (1 - \frac{1}{p_1})\cdot p_1^{q_1}(1 - \frac{1}{p_1})} \end{eqnarray*} $$

咱們令$t_1 < p_1$，即$p_1^{t_1}是gcd(a, b)$的一部分，那麼約分以後有： $$ \begin{eqnarray*} \frac{p_1^{t_1}}{p_1^{t_1} (1 - \frac{1}{p_1})} \end{eqnarray*} $$

咱們再同理考慮$p_1 \cdots p_n$，咱們發現分子恰好是$gcd(a, b)$， 而分母是$\varphi(gcd(a, b))$，即： $$ \begin{eqnarray*} \frac{\varphi(ab)}{\varphi(a)\varphi(b)} (\bmod p) &=& \frac{\varphi(p_1^{t_1 + q_1} \cdots p_n^{t_n + q_n})}{\varphi(p_1^{t_1} \cdots \varphi(p_n^{t_n})) \cdot \varphi(p_1^{q_1} \cdots p_n^{q_n})} (\bmod p) \ &=& \frac{gcd(a, b)}{\varphi(gcd(a, b))} \end{eqnarray*} $$

因此如今咱們的問題轉化成了求解： $$ \begin{eqnarray*} \sum\limits_{a = 1}^{m}\sum\limits_{b = 1}^{n} \frac{gcd(a, b)}{\varphi(gcd(a, b))} (\bmod p) \end{eqnarray*} $$

令$gcd(a, b) = d$，而且令$n <= m$，有： $$ \begin{eqnarray*} \sum\limits_{a = 1}^{m} \sum\limits_{b = 1}^{n} \frac{d}{\varphi(d)} = \sum\limits_{d = 1}^{n} d \cdot \varphi(d)^{-1} \sum\limits_{a = 1}^{n} \sum\limits_{b = 1}^{m} [gcd(a, b) == d] \ \end{eqnarray*} $$

咱們令： $$ \begin{eqnarray*} f(d) &=& \sum\limits_{a = 1}^{n} \sum\limits_{b = 1}^{m} [gcd(a, b) == d] \ g(d) &=& \sum\limits_{d|x}f(x) \ &=& \sum\limits_{a = 1}^{n} \sum\limits_{b = 1}^{m} [d | gcd(a, b)] \ &=& \sum\limits_{a = 1}^{n/d}\sum\limits_{b = 1}^{m/d} [1 | gcd(a, b)] \ &=& \lfloor \frac{n}{d} \rfloor \lfloor \frac{m}{d} \rfloor \end{eqnarray*} $$ 進行莫比烏斯反演，有： $$ \begin{eqnarray*} f(d) &=& \sum\limits_{d|x} \mu(\frac{x}{d}) g(d) \ &=& \sum\limits_{d|x} \mu(\frac{x}{d}) \cdot \lfloor \frac{n}{d} \rfloor \lfloor \frac{m}{d} \rfloor \ &=& \sum\limits_{x = 1}^{n/d} \mu(x) \cdot \lfloor \frac{n}{xd} \rfloor \lfloor \frac{m}{xd} \rfloor \ \end{eqnarray*} $$

因此，原式爲： $$ \begin{eqnarray*} \sum\limits_{i = 1}^{n} i \cdot \varphi(i)^{-1} \sum\limits_{d = 1}^{n|i} \mu(d) \lfloor \frac{n}{id} \rfloor \lfloor \frac{m}{id} \rfloor
\end{eqnarray*} $$ 預處理逆元，$\varphi()$函數，$\mu()$函數，而後直接算便可。 複雜度爲$\sum\limits_{i = 1}^{n} \sqrt{(i)}$

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

#define ll long long
#define N 1000010
ll p;
int n, m;
int prime[N], mu[N];
int phi[N], inv[N], g[N];
bool check[N];

void init()
{
	memset(check, 0, sizeof check);
	prime[0] = 0;
	phi[1] = 1;
	mu[1] = 1;
	for (int i = 2; i < N; ++i)
	{
		if (!check[i])
		{
			prime[++prime[0]] = i;
			phi[i] = i - 1;
			mu[i] = -1;
		}
		for (int j = 1; j <= prime[0]; ++j)
		{
			if (1ll * i * prime[j] >= N)
				break;
			check[i * prime[j]] = 1;
			if (i % prime[j] == 0)
			{
				phi[i * prime[j]] = phi[i] * prime[j];
				mu[i * prime[j]] = 0;
				break;
			}
			else
			{
				phi[i * prime[j]] = phi[i] * (prime[j] - 1);
				mu[i * prime[j]] = -mu[i]; 	
			}
		}
	}
}

void work()
{
	inv[1] = 1;
	for (int i = 2; i <= n; ++i)
		inv[i] = 1ll * inv[p % i] * (p - p / i) % p;
	for (int i = 1; i <= n; ++i)
		g[i] = 1ll * i * inv[phi[i]] % p;
}

ll get(int n, int m)
{
	ll res = 0;
	for (int i = 1; i <= n; ++i)
		res = (res + 1ll * mu[i] * (n / i) * (m / i)) % p;
	return res;
}

int main()
{
	init();
	int T; cin >> T;
	while (T--)
	{
		scanf("%d %d %lld\n", &n, &m, &p);
		if (n > m) swap(n, m);
		work();
		ll res = 0;
		for (int i = 1; i <= n; ++i)
			res = (res + g[i] * get(n / i, m / i)) % p; 
		printf("%lld\n", res);
	}
	return 0;
}