莫比乌斯相关

莫比乌斯函数

定义

μ (n) = {\begin{cases} 1 & n = 1 \\ 0 & n 含 有 平 方 因 子 \\ (- 1)^{k} & k 为 n 的 本 质 不 同 质 因 子 个 数 \end{cases}

$\mu(n) = \begin{cases} 1 & n = 1 \\ 0 & n 含有平方因子 \\ (-1)^k & k 为 n 的本质不同质因子个数\end{cases}$

性质

Dirichlet 卷积相关

\sum_{d | n} μ (d) = [n = 1] φ (n) = \sum_{d | n} μ (d) \times \frac{n}{d}

$\sum_{d | n} \mu(d) = [n = 1] \\ \varphi(n) = \sum_{d | n} \mu(d) \times \dfrac{n}{d}$

平方相关

μ^{2} (n) = \sum_{d^{2} | n} μ (d) \sum_{i = 1}^{n} μ^{2} (i) = \sum_{i = 1}^{n} \sum_{d^{2} | i} μ (d) = \sum_{d = 1}^{\sqrt{n}} μ (d) ⌊ \frac{n}{d^{2}} ⌋

$\mu^2(n) = \sum_{d^2 | n} \mu(d) \\ \sum_{i = 1}^n \mu^2(i) = \sum_{i = 1}^n \sum_{d^2 | i} \mu(d) = \sum_{d = 1}^{\sqrt{n}} \mu(d) \lfloor \dfrac{n}{d^2} \rfloor$

证明：考虑若 $\mu(i)^2=1$ ，则右侧只有一项 $j=1$ （因为 $i$ 无平方因子）否则说明其至少有一个质因子 $p$ 出现了两次以上。

考虑将所有 $j^2\mid i$ 分成两类： $p$ 是 $j$ 的因子/不是 $j$ 的因子（由于 $\mu(j)$ 不能为 $0$ ，因此 $p$ 至多出现一次）。则两类 $j$ 之间建立起一一对应关系，且对于每一对，它们的 $\mu$ 互为相反数。因此这个 $i$ 的贡献确实为 $0$ 。

求法

单个数

 inline int Mu(int n) {
	int mu = 1;
	
	for (int i = 2; i * i <= n; ++i)
		if (!(n % i)) {
			if (!(n % (i * i)))
				return 0;
 
			mu *= -1, n /= i;
		}
	
	if (n > 1)
		mu *= -1;
	
	return mu;
}

线性筛

 inline void sieve(int n) {
    memset(isp, true, sizeof(isp));
    isp[1] = false, mu[1] = 1;
    
    for (int i = 2; i <= n; ++i) {
        if (isp[i])
            pri[++pcnt] = i, mu[i] = -1;
        
        for (int j = 1; j <= pcnt && i * pri[j] <= n; ++j) {
            isp[i * pri[j]] = false;
            
            if (i % pri[j])
                mu[i * pri[j]] = -mu[i];
            else {
                mu[i * pri[j]] = 0;
                break;
            }
        }
    }
}

杜教筛

 namespace Mu {
map<int, ll> mp;
 
ll f[N], sum[N];
 
inline void prework() {
    for (int i = 1; i < N; ++i)
        sum[i] = sum[i - 1] + f[i];
}
 
ll Sum(ll n) {
    if (n < N)
        return sum[n];
 
    if (mp.find(n) != mp.end())
        return mp[n];
 
    ll res = 1;
 
    for (ll l = 2, r; l <= n; l = r + 1) {
        r = n / (n / l);
        res -= 1ll * (r - l + 1) * Sum(n / l);
    }
 
    return mp[n] = res;
}
} // namespace Mu

莫比乌斯反演

设 $f(n), g(n)$ 为两个数论函数，由 $\mu * 1 = \epsilon$ ，得到：

f = f * ϵ = f * (μ * 1) = (f * 1) * μ

$f = f * \epsilon = f * (\mu * 1) = (f * 1) * \mu$

形式一

设 $f(n), g(n)$ 为两个数论函数，则：

f (n) = \sum_{d | n} g (d) ⟺ g (n) = \sum_{d | n} f (d) μ (\frac{n}{d})

$f(n) = \sum_{d | n} g(d) \iff g(n) = \sum_{d | n} f(d) \mu(\dfrac{n}{d})$

此时 $f(n)$ 称为 $g(n)$ 的莫比乌斯变换， $g(n)$ 称为 $f(n)$ 的莫比乌斯逆变换，即莫比乌斯反演。

形式二

设 $f(n), g(n)$ 为两个数论函数，则：

f (n) = \sum_{n | d} g (d) ⟺ g (n) = \sum_{n | d} f (d) μ (\frac{d}{n})

$f(n) = \sum_{n | d} g(d) \iff g(n) = \sum_{n | d} f(d) \mu(\dfrac{d}{n})$

常见应用

形式一

\begin{aligned} (1) & \sum_{i = 1}^{n} \sum_{j = 1}^{m} [gcd (i, j) = 1] \\ (2) & = & \sum_{i = 1}^{n} \sum_{j = 1}^{m} \sum_{d | i} \sum_{d | j} μ (d) \\ (3) & = & \sum_{d = 1}^{n} μ (d) \sum_{i = 1}^{n} \sum_{j = 1}^{m} [d | i] [d | j] \\ (4) & = & \sum_{d = 1}^{n} μ (d) ⌊ \frac{n}{d} ⌋ ⌊ \frac{m}{d} ⌋ \end{aligned}

$\begin{align} &\sum_{i = 1}^n \sum_{j = 1}^m [\gcd(i, j) = 1] \\ =& \sum_{i = 1}^n \sum_{j = 1}^m \sum_{d | i} \sum_{d | j} \mu(d) \\ =& \sum_{d = 1}^n \mu(d) \sum_{i = 1}^n \sum_{j = 1}^m [d | i] [d | j] \\ =& \sum_{d = 1}^n \mu(d) \lfloor \dfrac{n}{d} \rfloor \lfloor \dfrac{m}{d} \rfloor \end{align}$

数论分块即可做到 $O(n) - O(\sqrt{n})$ 。

形式二

\begin{aligned} (5) & \sum_{i = 1}^{n} \sum_{j = 1}^{m} f (gcd (i, j)) \\ (6) & = & \sum_{d = 1}^{n} f (d) \sum_{i = 1}^{n} \sum_{j = 1}^{m} [gcd (i, j) = d] \\ (7) & = & \sum_{d = 1}^{n} f (d) \sum_{i = 1}^{⌊ \frac{n}{d} ⌋} \sum_{j = 1}^{⌊ \frac{m}{d} ⌋} [gcd (i, j) = 1] \\ (8) & = & \sum_{d = 1}^{n} f (d) \sum_{k = 1}^{⌊ \frac{n}{d} ⌋} μ (k) ⌊ \frac{n}{d k} ⌋ ⌊ \frac{m}{d k} ⌋ \end{aligned}

$\begin{align} & \sum_{i = 1}^n \sum_{j = 1}^m f(\gcd(i, j)) \\ =& \sum_{d = 1}^n f(d) \sum_{i = 1}^{n} \sum_{j = 1}^{m} [\gcd(i, j) = d] \\ =& \sum_{d = 1}^n f(d) \sum_{i = 1}^{\lfloor \frac{n}{d} \rfloor} \sum_{j = 1}^{\lfloor \frac{m}{d} \rfloor} [\gcd(i, j) = 1] \\ =& \sum_{d = 1}^n f(d) \sum_{k = 1}^{\lfloor \frac{n}{d} \rfloor} \mu(k) \lfloor \dfrac{n}{d k} \rfloor \lfloor \dfrac{m}{d k} \rfloor \end{align}$

令 $d k = T$ ，则：

= \sum_{T = 1}^{n} ⌊ \frac{n}{T} ⌋ ⌊ \frac{m}{T} ⌋ \sum_{d | T} f (d) μ (\frac{T}{d})

$= \sum_{T = 1}^n \lfloor \dfrac{n}{T} \rfloor \lfloor \dfrac{m}{T} \rfloor \sum_{d | T} f(d) \mu(\dfrac{T}{d})$

数论分块即可做到 $O(n) - O(\sqrt{n})$ 。

扩展

对于数论函数 $f, g$ 与完全积性函数 $t$ 且满足 $t(1) = 1$ ，则：

f (n) = \sum_{i = 1}^{n} t (i) g (⌊ \frac{n}{i} ⌋) ⟺ g (n) = \sum_{i = 1}^{n} μ (i) t (i) f (⌊ \frac{n}{i} ⌋)

$f(n) = \sum_{i = 1}^n t(i) g(\lfloor \dfrac{n}{i} \rfloor) \iff g(n) = \sum_{i = 1}^n \mu(i) t(i) f(\lfloor \dfrac{n}{i} \rfloor)$

证明：

\begin{aligned} (9) & g (n) & = \sum_{i = 1}^{n} μ (i) t (i) f (⌊ \frac{n}{i} ⌋) \\ (10) & = \sum_{i = 1}^{n} μ (i) t (i) \sum_{j = 1}^{⌊ \frac{n}{i} ⌋} t (j) g (⌊ \frac{n}{i j} ⌋) \\ (11) & = \sum_{T = 1}^{n} \sum_{i = 1}^{n} μ (i) t (i) \sum_{j = 1}^{⌊ \frac{n}{i} ⌋} [i j = T] t (j) g (⌊ \frac{n}{T} ⌋) \\ (12) & = \sum_{T = 1}^{n} \sum_{i | T} μ (i) t (i) t (\frac{T}{i}) g (⌊ \frac{n}{T} ⌋) \\ (13) & = \sum_{T = 1}^{n} g (⌊ \frac{n}{T} ⌋) t (T) \sum_{i | T} μ (i) \\ (14) & = g (n) t (1) \end{aligned}

$\begin{align} g(n) &= \sum_{i = 1}^n \mu(i) t(i) f(\lfloor \dfrac{n}{i} \rfloor) \\ &= \sum_{i = 1}^n \mu(i) t(i) \sum_{j = 1}^{\lfloor \frac{n}{i} \rfloor} t(j) g(\lfloor \dfrac{n}{ij} \rfloor) \\ &= \sum_{T = 1}^n \sum_{i = 1}^n \mu(i) t(i) \sum_{j = 1}^{\lfloor \frac{n}{i} \rfloor} [ij = T] t(j) g(\lfloor \dfrac{n}{T} \rfloor) \\ &= \sum_{T = 1}^n \sum_{i | T} \mu(i) t(i) t(\dfrac{T}{i}) g(\lfloor \dfrac{n}{T} \rfloor) \\ &= \sum_{T = 1}^n g(\lfloor \dfrac{n}{T} \rfloor) t(T) \sum_{i | T} \mu(i) \\ &= g(n) t(1) \end{align}$

魔力筛

这是一个可以在 $O(n\log\log n)$ 的时间复杂度求出任意数论函数与 $\mu$ 的狄利克雷卷积的算法。当然，必须保证这个数论函数能被提前计算出来。

设 $\mathbf g_{i,n}=\sum_{d\mid n}\mathbf f(d)\mu(\dfrac nd)$ ，其中 $d$ 只含前 $i$ 种质因子。则有：

g_{i, n} = {\begin{cases} g_{i - 1, n} & p_{i} ∤ n \\ g_{i - 1, n} - g_{i - 1, \frac{n}{p_{i}}} & p_{i} ∣ n \end{cases}

$\mathbf g_{i,n}= \begin{cases} \mathbf g_{i-1,n}&p_i\nmid n\\ \mathbf g_{i-1,n}-\mathbf g_{i-1,\frac n{p_i}}&p_i\mid n \end{cases}$

需要滚动数组优化。

解释：第一种情况显然正确。对于第二种，由于每多一个质因子， $\mu$ 的取值就会乘以 $-1$ ，因此上式的意义是强制不选 $p_i$ ，然后乘以 $-1$ 累加在答案上。如果含有平方因子，则值为 $0$ ，不影响答案。

复杂度和埃氏筛一样，为 $O(n \log \log n)$ 。

 inline void calc(int n) {
    for (int i = 1; i <= n; ++i)
        g[i] = f[i];
    
    for (int p : prime)
        for (int i = n / p; i; --i)
            g[i * p] = g[i * p] - g[i];
}

应用

P2257 YY的GCD PGCD - Primes in GCD Table

求：

$\sum_{i = 1}^n \sum_{j = 1}^m [\gcd(i, j) \in prime]$
$n, m \leq 10^7$

\begin{aligned} \sum_{i = 1}^{n} \sum_{j = 1}^{m} [gcd (i, j) \in p r i m e] \\ = & \sum_{k = 1}^{n} [k \in p r i m e] \sum_{i = 1}^{n} \sum_{j = 1}^{m} [gcd (i, j) = k] \\ = & \sum_{k = 1}^{n} [k \in p r i m e] \sum_{i = 1}^{⌊ \frac{n}{k} ⌋} \sum_{j = 1}^{⌊ \frac{m}{k} ⌋} [gcd (i, j) = 1] \\ = & \sum_{k = 1}^{n} [k \in p r i m e] \sum_{d = 1}^{⌊ \frac{n}{k} ⌋} ⌊ \frac{n}{d k} ⌋ ⌊ \frac{m}{d k} ⌋ μ (d) \end{aligned}

$\begin{aligned} &\sum_{i = 1}^n \sum_{j = 1}^m [\gcd(i, j) \in prime] \\ =& \sum_{k = 1}^n [k \in prime] \sum_{i = 1}^n \sum_{j = 1}^m [\gcd(i, j) = k] \\ =& \sum_{k = 1}^n [k \in prime] \sum_{i = 1}^{\lfloor \frac{n}{k} \rfloor} \sum_{j = 1}^{\lfloor \frac{m}{k} \rfloor} [\gcd(i, j) = 1] \\ =& \sum_{k = 1}^n [k \in prime] \sum_{d = 1}^{\lfloor \frac{n}{k} \rfloor} \lfloor \dfrac{n}{dk} \rfloor \lfloor \dfrac{m}{dk} \rfloor \mu(d) \end{aligned}$

设 $T = dk$ ，则：

\begin{aligned} = \sum_{k = 1}^{n} [k \in p r i m e] \sum_{d = 1}^{⌊ \frac{n}{k} ⌋} ⌊ \frac{n}{T} ⌋ ⌊ \frac{m}{T} ⌋ μ (d) \\ = \sum_{T = 1}^{n} ⌊ \frac{n}{T} ⌋ ⌊ \frac{m}{T} ⌋ \sum_{k | T, k \in p r i m e} μ (\frac{T}{k}) \end{aligned}

$\begin{aligned} &= \sum_{k = 1}^n [k \in prime] \sum_{d = 1}^{\lfloor \frac{n}{k} \rfloor} \lfloor \dfrac{n}{T} \rfloor \lfloor \dfrac{m}{T} \rfloor \mu(d) \\ &= \sum_{T = 1}^n \lfloor \dfrac{n}{T} \rfloor \lfloor \dfrac{m}{T} \rfloor \sum_{k | T, k \in prime} \mu(\dfrac{T}{k}) \end{aligned}$

$n = m$ 的弱化版是这题：P2568 GCD

 #include <bits/stdc++.h>
typedef long long ll;
using namespace std;
const int N = 1e7 + 7;
 
ll f[N], sum[N];
int pri[N], mu[N];
bool isp[N];
 
int T, n, m, pcnt;
 
inline void sieve() {
	memset(isp, true, sizeof(isp));
    mu[1] = 1, isp[1] = false;
 
    for (int i = 2; i < N; ++i) {
        if (isp[i])
            pri[++pcnt] = i, mu[i] = -1;
 
        for (int j = 1; j <= pcnt && i * pri[j] < N; ++j) {
            isp[pri[j] * i] = false;
 
            if (i % pri[j])
                mu[i * pri[j]] = -mu[i];
            else
                break;
        }
    }
    
    for (int i = 1; i <= pcnt; ++i)
    	for (int j = 1; pri[i] * j < N; ++j)
    		f[j * pri[i]] += mu[j];
 
    for (int i = 1; i < N; ++i)
        sum[i] = sum[i - 1] + f[i];
}
 
signed main() {
    sieve();
    scanf("%d", &T);
 
    while (T--) {
        scanf("%d%d", &n, &m);
        
        if (n > m)
        	swap(n, m);
        
        ll ans = 0;
 
        for (int l = 1, r = 0; l <= n; l = r + 1) {
        	r = min(n / (n / l), m / (m / l));
        	ans += (sum[r] - sum[l - 1]) * (n / l) * (m / l);
        }
 
        printf("%lld\n", ans);
    }
 
    return 0;
}

P3327 [SDOI2015] 约数个数和

设 $d(n)$ 表示 $n$ 的约数个数，求：

$\sum_{i = 1}^n \sum_{j = 1}^m d(ij)$
$T, n, m \leq 5 \times 10^4$

\begin{aligned} (15) & \sum_{i = 1}^{n} \sum_{j = 1}^{m} d (i j) \\ (16) & = & \sum_{i = 1}^{n} \sum_{j = 1}^{m} \sum_{x | i} \sum_{y | j} [gcd (x, y) = 1] \\ (17) & = & \sum_{x = 1}^{n} \sum_{y = 1}^{m} ⌊ \frac{n}{x} ⌋ ⌊ \frac{m}{y} ⌋ \sum_{d | x, d | y} μ (d) \\ (18) & = & \sum_{d = 1}^{n} μ (d) \sum_{i = 1}^{⌊ \frac{n}{d} ⌋} \sum_{j = 1}^{⌊ \frac{m}{d} ⌋} ⌊ \frac{n}{i d} ⌋ ⌊ \frac{m}{j d} ⌋ \\ (19) & = & \sum_{d = 1}^{n} μ (d) \sum_{i = 1}^{⌊ \frac{n}{d} ⌋} ⌊ \frac{n}{i d} ⌋ \sum_{j = 1}^{⌊ \frac{m}{d} ⌋} ⌊ \frac{m}{j d} ⌋ \end{aligned}

$\begin{align} & \sum_{i = 1}^n \sum_{j = 1}^m d(ij) \\ =& \sum_{i = 1}^n \sum_{j = 1}^m \sum_{x | i} \sum_{y | j} [\gcd(x, y) = 1] \\ =& \sum_{x = 1}^n \sum_{y = 1}^m \lfloor \dfrac{n}{x} \rfloor \lfloor \dfrac{m}{y} \rfloor \sum_{d|x, d | y} \mu(d) \\ =& \sum_{d = 1}^n \mu(d) \sum_{i = 1}^{\lfloor \frac{n}{d} \rfloor} \sum_{j = 1}^{\lfloor \frac{m}{d} \rfloor} \lfloor \dfrac{n}{id} \rfloor \lfloor \dfrac{m}{jd} \rfloor \\ =& \sum_{d = 1}^n \mu(d) \sum_{i = 1}^{\lfloor \frac{n}{d} \rfloor} \lfloor \dfrac{n}{id} \rfloor \sum_{j = 1}^{\lfloor \frac{m}{d} \rfloor} \lfloor \dfrac{m}{jd} \rfloor \\ \end{align}$

 #include <bits/stdc++.h>
typedef long long ll;
using namespace std;
const int N = 5e4 + 7;
 
ll f[N];
int pri[N], mu[N], sum[N];
bool isp[N];
 
int T, n, m, pcnt;
 
inline void sieve() {
	memset(isp, true, sizeof(isp));
	isp[1] = false, mu[1] = 1;
 
	for (int i = 2; i < N; ++i) {
		if (isp[i])
			pri[++pcnt] = i, mu[i] = -1;
 
		for (int j = 1; j <= pcnt && i * pri[j] < N; ++j) {
			isp[i * pri[j]] = false;
 
			if (i % pri[j])
				mu[i * pri[j]] = -mu[i];
			else {
				mu[i * pri[j]] = 0;
				break;
			}
		}
	}
 
	for (int i = 1; i < N; ++i)
		sum[i] = sum[i - 1] + mu[i];
 
	for (int i = 1; i < N; ++i)
		for (int l = 1, r; l <= i; l = r + 1) {
			r = i / (i / l);
			f[i] += 1ll * (i / l) * (r - l + 1);
		}
}
 
signed main() {
	sieve();
	scanf("%d", &T);
 
	while (T--) {
		scanf("%d%d", &n, &m);
 
		if (n > m)
			swap(n, m);
 
		ll ans = 0;
 
		for (int l = 1, r; l <= n; l = r + 1) {
			r = min(n / (n / l), m / (m / l));
			ans += f[n / l] * f[m / l] * (sum[r] - sum[l - 1]);
		}
 
		printf("%lld\n", ans);
	}
 
	return 0;
}

P5221 Product

求：

$\prod_{i = 1}^n \prod_{j = 1}^n \dfrac{\operatorname{lcm}(i, j)}{\gcd(i, j)} \pmod{104857601}$
$n \leq 10^6$

\begin{aligned} (20) & \prod_{i = 1}^{n} \prod_{j = 1}^{n} \frac{lcm (i, j)}{gcd (i, j)} \\ (21) & = & \prod_{i = 1}^{n} \prod_{j = 1}^{n} \frac{i j}{gcd (i, j)^{2}} \\ (22) & = & \frac{\prod_{i = 1}^{n} i \prod_{j = 1}^{n} j}{\prod_{i = 1}^{n} \prod_{j = 1}^{n} gcd (i, j)^{2}} \\ (23) & = & \frac{(n!)^{2 n}}{(\prod_{i = 1}^{n} \prod_{j = 1}^{n} gcd (i, j))^{2}} \end{aligned}

$\begin{align} & \prod_{i = 1}^n \prod_{j = 1}^n \dfrac{\operatorname{lcm}(i, j)}{\gcd(i, j)} \\ =& \prod_{i = 1}^n \prod_{j = 1}^n \dfrac{ij}{\gcd(i, j)^2} \\ =& \dfrac{\prod_{i = 1}^n i \prod_{j = 1}^n j}{\prod_{i = 1}^n \prod_{j = 1}^n \gcd(i, j)^2} \\ =& \dfrac{(n!)^{2n}}{(\prod_{i = 1}^n \prod_{j = 1}^n \gcd(i, j))^2} \\ \end{align}$

单独考虑下面的部分得到：

\prod_{d = 1}^{n} d^{\sum_{i = 1}^{n} \sum_{j = 1}^{n} [gcd (i, j) = d]}

$\prod_{d = 1}^n d^{\sum_{i = 1}^{n} \sum_{j = 1}^{n} [\gcd(i, j) = d] \\}$

单独考虑指数部分得到：

\begin{aligned} (24) & \sum_{i = 1}^{⌊ \frac{n}{d} ⌋} \sum_{j = 1}^{⌊ \frac{n}{d} ⌋} [gcd (i, j) = 1] \\ (25) & = & (\sum_{i = 1}^{⌊ \frac{n}{d} ⌋} (2 \times \sum_{j = 1}^{i} [gcd (i, j) = 1])) - 1 \\ (26) & = & 2 \times (\sum_{i = 1}^{⌊ \frac{n}{d} ⌋} φ (i)) - 1 \end{aligned}

$\begin{align} & \sum_{i = 1}^{\lfloor \frac{n}{d} \rfloor} \sum_{j = 1}^{\lfloor \frac{n}{d} \rfloor} [\gcd(i, j) = 1] \\ =& \left( \sum_{i = 1}^{\lfloor \frac{n}{d} \rfloor} \left(2 \times \sum_{j = 1}^i [\gcd(i, j) = 1] \right) \right) - 1 \\ =& 2 \times \left( \sum_{i = 1}^{\lfloor \frac{n}{d} \rfloor} \varphi(i) \right) - 1 \\ \end{align}$

 #include <bits/stdc++.h>
typedef long long ll;
using namespace std;
const int Mod = 104857601;
const int N = 1e6 + 7;
 
int pri[N], phi[N], sum[N];
bool isp[N];
 
int n, pcnt;
 
inline int add(int x, int y) {
	x += y;
	
	if (x >= Mod)
		x -= Mod;
	
	return x;
}
 
inline int dec(int x, int y) {
	x -= y;
	
	if (x < 0)
		x += Mod;
	
	return x;
}
 
inline int mi(int a, int b) {
	int res = 1;
	
	for (; b; b >>= 1, a = 1ll * a * a % Mod)
		if (b & 1)
			res = 1ll * res * a % Mod;
	
	return res;
}
 
inline void sieve(int n) {
	memset(isp, true, sizeof(isp));
	isp[1] = false, phi[1] = 1;
 
	for (int i = 2; i <= n; ++i) {
		if (isp[i])
			pri[++pcnt] = i, phi[i] = i - 1;
 
		for (int j = 1; j <= pcnt && i * pri[j] < N; ++j) {
			isp[i * pri[j]] = false;
 
			if (i % pri[j])
				phi[i * pri[j]] = phi[i] * phi[pri[j]];
			else {
				phi[i * pri[j]] = phi[i] * pri[j];
				break;
			}
		}
	}
 
	for (int i = 1; i <= n; ++i)
		sum[i] = (sum[i - 1] + phi[i]) % (Mod - 1);
}
 
signed main() {
	scanf("%d", &n);
	sieve(n);
	int fac = 1;
 
	for (int i = 1; i <= n; ++i)
		fac = 1ll * fac * i % Mod;
 
	int ans = 1ll * mi(fac, n * 2) * fac % Mod * fac % Mod;
 
	for (int i = 1; i <= n; ++i)
		ans = 1ll * ans * mi(mi(i, 4ll * sum[n / i] % (Mod - 1)), Mod - 2) % Mod;
 
	printf("%d", ans);
	return 0;
}

P6055 [RC-02] GCD

求：

$\sum_{i = 1}^{n} \sum_{j = 1}^n \sum_{p = 1}^{\lfloor \frac{n}{j} \rfloor} \sum_{q = 1}^{\lfloor \frac{n}{j} \rfloor} [\gcd(i, j) = 1] [\gcd(p, q) = 1] \pmod{998244353}$
$n \leq 2 \times 10^9$

\begin{aligned} (27) & \sum_{i = 1}^{n} \sum_{j = 1}^{n} \sum_{p = 1}^{⌊ \frac{n}{j} ⌋} \sum_{q = 1}^{⌊ \frac{n}{j} ⌋} [gcd (i, j) = 1] [gcd (p, q) = 1] \\ (28) & = & \sum_{i = 1}^{n} \sum_{j = 1}^{n} [gcd (i, j) = 1] \sum_{p = 1}^{n} \sum_{q = 1}^{n} [gcd (p, q) = j] \\ (29) & = & \sum_{i = 1}^{n} \sum_{p = 1}^{n} \sum_{q = 1}^{n} [gcd (i, p, q) = 1] \\ (30) & = & \sum_{d = 1}^{n} μ (d) ⌊ \frac{n}{d} ⌋^{3} \end{aligned}

$\begin{align} & \sum_{i = 1}^{n} \sum_{j = 1}^n \sum_{p = 1}^{\lfloor \frac{n}{j} \rfloor} \sum_{q = 1}^{\lfloor \frac{n}{j} \rfloor} [\gcd(i, j) = 1] [\gcd(p, q) = 1] \\ =& \sum_{i = 1}^{n} \sum_{j = 1}^n [\gcd(i, j) = 1] \sum_{p = 1}^{n} \sum_{q = 1}^{n} [\gcd(p, q) = j] \\ =& \sum_{i = 1}^{n} \sum_{p = 1}^{n} \sum_{q = 1}^{n} [\gcd(i, p, q) = 1] \\ =& \sum_{d = 1}^n \mu(d) \lfloor \dfrac{n}{d} \rfloor^3 \end{align}$

杜教筛处理 $\mu$ 的前缀和即可。

 #include <bits/stdc++.h>
using namespace std;
const int Mod = 998244353;
const int N = 1e7 + 7;
 
map<int, int> mp;
 
int pri[N], mu[N], sum[N];
bool isp[N];
 
int n, pcnt;
 
inline int add(int x, int y) {
	x += y;
	
	if (x >= Mod)
		x -= Mod;
	
	return x;
}
 
inline int dec(int x, int y) {
	x -= y;
	
	if (x < 0)
		x += Mod;
	
	return x;
}
 
inline void sieve() {
	memset(isp, true, sizeof(isp));
	isp[1] = false, mu[1] = 1;
 
	for (int i = 2; i < N; ++i) {
		if (isp[i])
			pri[++pcnt] = i, mu[i] = Mod - 1;
 
		for (int j = 1; j <= pcnt && i * pri[j] < N; ++j) {
			isp[i * pri[j]] = false;
 
			if (i % pri[j])
				mu[i * pri[j]] = Mod - mu[i];
			else {
				mu[i * pri[j]] = 0;
				break;
			}
		}
	}
 
	for (int i = 1; i < N; ++i)
		sum[i] = add(sum[i - 1], mu[i]);
}
 
int Sum(int n) {
	if (n < N)
		return sum[n];
 
	if (mp.find(n) != mp.end())
		return mp[n];
 
	int res = 1;
 
	for (int l = 2, r; l <= n; l = r + 1) {
		r = n / (n / l);
		res = dec(res, 1ll * (r - l + 1) * Sum(n / l) % Mod);
	}
 
	return mp[n] = res;
}
 
signed main() {
	sieve();
	scanf("%d", &n);
	int ans = 0;
 
	for (int l = 1, r; l <= n; l = r + 1) {
		r = n / (n / l);
		ans = add(ans, 1ll * (n / l) * (n / l) % Mod * (n / l) % Mod * dec(Sum(r), Sum(l - 1)) % Mod);
	}
 
	printf("%d", ans);
	return 0;
}

P4449 于神之怒加强版

求：

$\sum_{i = 1}^n \sum_{j = 1}^m \gcd(i, j)^k \pmod{10^9 + 7}$
$n, m, k \leq 5 \times 10^6$

\begin{aligned} (31) & \sum_{i = 1}^{n} \sum_{j = 1}^{m} gcd (i, j)^{k} \\ (32) & = & \sum_{d = 1}^{n} d^{k} \sum_{i = 1}^{⌊ \frac{n}{d} ⌋} \sum_{j = 1}^{⌊ \frac{m}{d} ⌋} [gcd (i, j) = 1] \\ (33) & = & \sum_{d = 1}^{n} d^{k} \sum_{k = 1}^{⌊ \frac{n}{d} ⌋} μ (k) ⌊ \frac{n}{d k} ⌋ ⌊ \frac{m}{d k} ⌋ \\ (34) & = & \sum_{T = 1}^{n} ⌊ \frac{n}{T} ⌋ ⌊ \frac{m}{T} ⌋ \sum_{d | T} d^{k} μ (\frac{T}{d}) \end{aligned}

$\begin{align} & \sum_{i = 1}^n \sum_{j = 1}^m \gcd(i, j)^k \\ =& \sum_{d = 1}^n d^k \sum_{i = 1}^{\lfloor \frac{n}{d} \rfloor} \sum_{j = 1}^{\lfloor \frac{m}{d} \rfloor} [\gcd(i, j) = 1] \\ =& \sum_{d = 1}^n d^k \sum_{k = 1}^{\lfloor \frac{n}{d} \rfloor} \mu(k) \lfloor \dfrac{n}{dk} \rfloor \lfloor \dfrac{m}{dk} \rfloor \\ =& \sum_{T = 1}^n \lfloor \dfrac{n}{T} \rfloor \lfloor \dfrac{m}{T} \rfloor \sum_{d | T} d^k \mu(\dfrac{T}{d}) \end{align}$

线性筛 $g(T) = \sum_{d | T} d^k \mu(\dfrac{T}{d})$ 即可。

 #include <bits/stdc++.h>
using namespace std;
const int Mod = 1e9 + 7;
const int N = 5e6 + 7;
 
int pri[N], low[N], g[N], sum[N];
bool isp[N];
 
int T, n, m, k, pcnt;
 
inline int add(int x, int y) {
	x += y;
	
	if (x >= Mod)
		x -= Mod;
	
	return x;
}
 
inline int dec(int x, int y) {
	x -= y;
	
	if (x < 0)
		x += Mod;
	
	return x;
}
 
inline int mi(int a, int b) {
	int res = 1;
	
	for (; b; b >>= 1, a = 1ll * a * a % Mod)
		if (b & 1)
			res = 1ll * res * a % Mod;
	
	return res;
}
 
inline void sieve() {
	memset(isp, true, sizeof(isp));
	isp[1] = false, low[1] = 1, g[1] = 1;
 
	for (int i = 2; i < N; ++i) {
		if (isp[i]) {
			pri[++pcnt] = i, low[i] = i, g[i] = dec(mi(i, k), 1);
			int x = i;
 
			while (1ll * x * i < N)
				x *= i, low[x] = x, g[x] = dec(mi(x, k), mi(x / i, k));
		}
 
		for (int j = 1; j <= pcnt && i * pri[j] < N; ++j) {
			int x = i * pri[j];
			isp[x] = false;
			low[x] = i % pri[j] ? pri[j] : low[i] * pri[j];
			g[x] = 1ll * g[x / low[x]] * g[low[x]] % Mod;
 
			if (!(i % pri[j]))
				break;
		}
	}
 
	for (int i = 1; i < N; ++i)
		sum[i] = add(sum[i - 1], g[i]);
}
 
signed main() {
	scanf("%d%d", &T, &k);
	sieve();
 
	while (T--) {
		scanf("%d%d", &n, &m);
		int ans = 0;
 
		if (n > m)
			swap(n, m);
 
		for (int l = 1, r; l <= n; l = r + 1) {
			r = min(n / (n / l), m / (m / l));
			ans = add(ans, 1ll * (n / l) * (m / l) % Mod * dec(sum[r], sum[l - 1]) % Mod);
		}
 
		printf("%d\n", ans);
	}
 
	return 0;
}

P3768 简单的数学题

求：

$\sum_{i = 1}^n \sum_{j = 1}^n ij \gcd(i, j) \pmod{p}$
$n \leq 10^{10}$

\begin{aligned} (35) & \sum_{i = 1}^{n} \sum_{j = 1}^{n} i j gcd (i, j) \\ (36) & = & \sum_{d = 1}^{n} \sum_{i = 1}^{⌊ \frac{n}{d} ⌋} \sum_{j = 1}^{⌊ \frac{n}{d} ⌋} d \times i d \times j d \times [gcd (i, j) = 1] \\ (37) & = & \sum_{d = 1}^{n} d^{3} \sum_{i = 1}^{⌊ \frac{n}{d} ⌋} \sum_{j = 1}^{⌊ \frac{n}{d} ⌋} i j \sum_{k | i, k | j} μ (k) \\ (38) & = & \sum_{d = 1}^{n} d^{3} \sum_{k = 1}^{⌊ \frac{n}{d} ⌋} μ (k) \sum_{i = 1}^{⌊ \frac{n}{d k} ⌋} \sum_{j = 1}^{⌊ \frac{n}{d k} ⌋} i j k^{2} \\ (39) & = & \sum_{d = 1}^{n} d^{3} \sum_{k = 1}^{⌊ \frac{n}{d} ⌋} μ (k) k^{2} S (⌊ \frac{n}{d k} ⌋)^{2} \end{aligned}

$\begin{align} & \sum_{i = 1}^n \sum_{j = 1}^n ij \gcd(i, j) \\ =& \sum_{d = 1}^n \sum_{i = 1}^{\lfloor \frac{n}{d} \rfloor} \sum_{j = 1}^{\lfloor \frac{n}{d} \rfloor} d \times id \times jd \times [\gcd(i, j) = 1] \\ =& \sum_{d = 1}^n d^3 \sum_{i = 1}^{\lfloor \frac{n}{d} \rfloor} \sum_{j = 1}^{\lfloor \frac{n}{d} \rfloor} ij \sum_{k | i, k | j} \mu(k) \\ =& \sum_{d = 1}^n d^3 \sum_{k = 1}^{\lfloor \frac{n}{d} \rfloor} \mu(k) \sum_{i = 1}^{\lfloor \frac{n}{dk} \rfloor} \sum_{j = 1}^{\lfloor \frac{n}{dk} \rfloor} ijk^2 \\ =& \sum_{d = 1}^n d^3 \sum_{k = 1}^{\lfloor \frac{n}{d} \rfloor} \mu(k) k^2 S(\lfloor \dfrac{n}{dk} \rfloor)^2 \end{align}$

其中 $S(n) = \dfrac{n(n + 1)}{2}$ ，令 $T = dk$ 则：

\begin{aligned} (40) & = & \sum_{T = 1}^{n} T^{2} S (⌊ \frac{n}{T} ⌋)^{2} \sum_{d | T} d μ (\frac{T}{d}) \\ (41) & = & \sum_{T = 1}^{n} S (⌊ \frac{n}{T} ⌋)^{2} T^{2} φ (T) \end{aligned}

$\begin{align} =& \sum_{T = 1}^n T^2 S(\lfloor \dfrac{n}{T} \rfloor)^2 \sum_{d | T} d \mu(\dfrac{T}{d}) \\ =& \sum_{T = 1}^n S(\lfloor \dfrac{n}{T} \rfloor)^2 T^2 \varphi(T) \end{align}$

杜教筛处理 $T^2 \varphi(T)$ 的前缀和即可，具体为构造 $g(n) = n^2$ 。

 #include <bits/stdc++.h>
typedef long long ll;
using namespace std;
const int N = 1e7 + 7;
 
map<ll, int> mp;
 
int pri[N], phi[N], sum[N];
bool isp[N];
 
ll n;
int Mod, pcnt, inv6;
 
inline int add(int x, int y) {
	x += y;
	
	if (x >= Mod)
		x -= Mod;
	
	return x;
}
 
inline int dec(int x, int y) {
	x -= y;
	
	if (x < 0)
		x += Mod;
	
	return x;
}
 
inline int mi(int a, int b) {
	int res = 1;
	
	for (; b; b >>= 1, a = 1ll * a * a % Mod)
		if (b & 1)
			res = 1ll * res * a % Mod;
	
	return res;
}
 
inline void sieve() {
	memset(isp, true, sizeof(isp));
	isp[1] = false, phi[1] = 1;
 
	for (int i = 2; i < N; ++i) {
		if (isp[i])
			pri[++pcnt] = i, phi[i] = i - 1;
 
		for (int j = 1; j <= pcnt && i * pri[j] < N; ++j) {
			isp[i * pri[j]] = false;
 
			if (i % pri[j])
				phi[i * pri[j]] = phi[i] * phi[pri[j]];
			else {
				phi[i * pri[j]] = phi[i] * pri[j];
				break;
			}
		}
	}
 
	for (int i = 1; i < N; ++i)
		sum[i] = add(sum[i - 1], 1ll * i * i % Mod * phi[i] % Mod);
}
 
inline int Sum1(ll n) {
	n %= Mod;
	return n * (n + 1) / 2 % Mod;
}
 
inline int Sum2(ll n) {
	n %= Mod;
	return n * (n + 1) % Mod * (n * 2 + 1) % Mod * inv6 % Mod;
}
 
int Sum(ll n) {
	if (n < N)
		return sum[n];
 
	if (mp.find(n) != mp.end())
		return mp[n];
 
	int res = 1ll * Sum1(n) * Sum1(n) % Mod;
 
	for (ll l = 2, r; l <= n; l = r + 1) {
		r = n / (n / l);
		res = dec(res, 1ll * dec(Sum2(r), Sum2(l - 1)) * Sum(n / l) % Mod);
	}
 
	return mp[n] = res;
}
 
signed main() {
	scanf("%d%lld", &Mod, &n);
	sieve();
	inv6 = mi(6, Mod - 2);
	int ans = 0;
 
	for (ll l = 1, r; l <= n; l = r + 1) {
		r = n / (n / l);
		ans = add(ans, 1ll * Sum1(n / l) * Sum1(n / l) % Mod * dec(Sum(r), Sum(l - 1)) % Mod);
	}
 
	printf("%d", ans);
	return 0;
}

P3172 [CQOI2015] 选数

求：

$\sum_{i_1 = L}^R \sum_{i_2 = L}^R \cdots \sum_{i_n = L}^R [\gcd(i_1, i_2, \cdots, i_n) = k] \pmod{10^9 + 7}$
$n, k \leq 10^9$ ， $L, R \leq 10^9$ ， $R - L \leq 10^5$

令 $l = \lceil \dfrac{L}{k} \rceil, r = \lfloor \dfrac{R}{k} \rfloor$ ，则：

\begin{aligned} (42) & \sum_{i_{1} = L}^{R} \sum_{i_{2} = L}^{R} \dots \sum_{i_{n} = L}^{R} [gcd (i_{1}, i_{2}, \dots, i_{n}) = k] \\ (43) & = & \sum_{i_{1} = l}^{r} \sum_{i_{2} = l}^{r} \dots \sum_{i_{n} = l}^{r} [gcd (i_{1}, i_{2}, \dots, i_{n}) = 1] \\ (44) & = & \sum_{i_{1} = l}^{r} \sum_{i_{2} = l}^{r} \dots \sum_{i_{n} = l}^{r} \sum_{d | i_{1}, d | i_{2}, \dots, d | i_{n}} μ (d) \\ (45) & = & \sum_{d = 1}^{r} μ (d) (⌊ \frac{r}{d} ⌋ - ⌊ \frac{l - 1}{d} ⌋)^{n} \end{aligned}

$\begin{align} & \sum_{i_1 = L}^R \sum_{i_2 = L}^R \cdots \sum_{i_n = L}^R [\gcd(i_1, i_2, \cdots, i_n) = k] \\ =& \sum_{i_1 = l}^r \sum_{i_2 = l}^r \cdots \sum_{i_n = l}^r [\gcd(i_1, i_2, \cdots, i_n) = 1] \\ =& \sum_{i_1 = l}^r \sum_{i_2 = l}^r \cdots \sum_{i_n = l}^r \sum_{d | i_1, d | i_2, \cdots, d | i_n} \mu(d) \\ =& \sum_{d = 1}^r \mu(d) (\lfloor \dfrac{r}{d} \rfloor - \lfloor \dfrac{l - 1}{d} \rfloor)^n \end{align}$

 #include <bits/stdc++.h>
using namespace std;
const int Mod = 1e9 + 7;
const int N = 1e7 + 7;
 
map<int, int> mp;
 
int pri[N], mu[N], sum[N];
bool isp[N];
 
int n, k, L, R, pcnt;
 
inline int add(int x, int y) {
	x += y;
	
	if (x >= Mod)
		x -= Mod;
	
	return x;
}
 
inline int dec(int x, int y) {
	x -= y;
	
	if (x < 0)
		x += Mod;
	
	return x;
}
 
inline int mi(int a, int b) {
	int res = 1;
	
	for (; b; b >>= 1, a = 1ll * a * a % Mod)
		if (b & 1)
			res = 1ll * res * a % Mod;
	
	return res;
}
 
inline void sieve() {
	memset(isp, true, sizeof(isp));
	isp[1] = false, mu[1] = 1;
 
	for (int i = 2; i < N; ++i) {
		if (isp[i])
			pri[++pcnt] = i, mu[i] = Mod - 1;
 
		for (int j = 1; j <= pcnt && i * pri[j] < N; ++j) {
			isp[i * pri[j]] = false;
 
			if (i % pri[j])
				mu[i * pri[j]] = Mod - mu[i];
			else {
				mu[i * pri[j]] = 0;
				break;
			}
		}
	}
 
	for (int i = 1; i < N; ++i)
		sum[i] = add(sum[i - 1], mu[i]);
}
 
inline int Sum(int n) {
	if (n < N)
		return sum[n];
 
	if (mp.find(n) != mp.end())
		return mp[n];
 
	int res = 1;
 
	for (int l = 2, r; l <= n; l = r + 1) {
		r = n / (n / l);
		res = dec(res, 1ll * (r - l + 1) * Sum(n / l) % Mod);
	}
 
	return mp[n] = res;
}
 
signed main() {
	sieve();
	scanf("%d%d%d%d", &n, &k, &L, &R);
	L = (L - 1) / k, R /= k;
	int ans = 0;
 
	for (int l = 1, r; l <= R; l = r + 1) {
		r = R / (R / l);
 
		if (L / l)
			r = min(r, L / (L / l));
 
		ans = add(ans, 1ll * mi(R / l - L / l, n) * dec(Sum(r), Sum(l - 1)) % Mod);
	}
 
	printf("%d", ans);
	return 0;
}

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	inline int Mu(int n) {
	int mu = 1;

	for (int i = 2; i * i <= n; ++i)
	if (!(n % i)) {
	if (!(n % (i * i)))
	return 0;

	mu *= -1, n /= i;
	}

	if (n > 1)
	mu *= -1;

	return mu;
	}

	inline void sieve(int n) {
	memset(isp, true, sizeof(isp));
	isp[1] = false, mu[1] = 1;

	for (int i = 2; i <= n; ++i) {
	if (isp[i])
	pri[++pcnt] = i, mu[i] = -1;

	for (int j = 1; j <= pcnt && i * pri[j] <= n; ++j) {
	isp[i * pri[j]] = false;

	if (i % pri[j])
	mu[i * pri[j]] = -mu[i];
	else {
	mu[i * pri[j]] = 0;
	break;
	}
	}
	}
	}

	namespace Mu {
	map<int, ll> mp;

	ll f[N], sum[N];

	inline void prework() {
	for (int i = 1; i < N; ++i)
	sum[i] = sum[i - 1] + f[i];
	}

	ll Sum(ll n) {
	if (n < N)
	return sum[n];

	if (mp.find(n) != mp.end())
	return mp[n];

	ll res = 1;

	for (ll l = 2, r; l <= n; l = r + 1) {
	r = n / (n / l);
	res -= 1ll * (r - l + 1) * Sum(n / l);
	}

	return mp[n] = res;
	}
	} // namespace Mu

	inline void calc(int n) {
	for (int i = 1; i <= n; ++i)
	g[i] = f[i];

	for (int p : prime)
	for (int i = n / p; i; --i)
	g[i * p] = g[i * p] - g[i];
	}

	#include <bits/stdc++.h>
	typedef long long ll;
	using namespace std;
	const int N = 1e7 + 7;

	ll f[N], sum[N];
	int pri[N], mu[N];
	bool isp[N];

	int T, n, m, pcnt;

	inline void sieve() {
	memset(isp, true, sizeof(isp));
	mu[1] = 1, isp[1] = false;

	for (int i = 2; i < N; ++i) {
	if (isp[i])
	pri[++pcnt] = i, mu[i] = -1;

	for (int j = 1; j <= pcnt && i * pri[j] < N; ++j) {
	isp[pri[j] * i] = false;

	if (i % pri[j])
	mu[i * pri[j]] = -mu[i];
	else
	break;
	}
	}

	for (int i = 1; i <= pcnt; ++i)
	for (int j = 1; pri[i] * j < N; ++j)
	f[j * pri[i]] += mu[j];

	for (int i = 1; i < N; ++i)
	sum[i] = sum[i - 1] + f[i];
	}

	signed main() {
	sieve();
	scanf("%d", &T);

	while (T--) {
	scanf("%d%d", &n, &m);

	if (n > m)
	swap(n, m);

	ll ans = 0;

	for (int l = 1, r = 0; l <= n; l = r + 1) {
	r = min(n / (n / l), m / (m / l));
	ans += (sum[r] - sum[l - 1]) * (n / l) * (m / l);
	}

	printf("%lld\n", ans);
	}

	return 0;
	}

	#include <bits/stdc++.h>
	using namespace std;
	const int Mod = 998244353;
	const int N = 1e7 + 7;

	map<int, int> mp;

	int pri[N], mu[N], sum[N];
	bool isp[N];

	int n, pcnt;

	inline int add(int x, int y) {
	x += y;

	if (x >= Mod)
	x -= Mod;

	return x;
	}

	inline int dec(int x, int y) {
	x -= y;

	if (x < 0)
	x += Mod;

	return x;
	}

	inline void sieve() {
	memset(isp, true, sizeof(isp));
	isp[1] = false, mu[1] = 1;

	for (int i = 2; i < N; ++i) {
	if (isp[i])
	pri[++pcnt] = i, mu[i] = Mod - 1;

	for (int j = 1; j <= pcnt && i * pri[j] < N; ++j) {
	isp[i * pri[j]] = false;

	if (i % pri[j])
	mu[i * pri[j]] = Mod - mu[i];
	else {
	mu[i * pri[j]] = 0;
	break;
	}
	}
	}

	for (int i = 1; i < N; ++i)
	sum[i] = add(sum[i - 1], mu[i]);
	}

	int Sum(int n) {
	if (n < N)
	return sum[n];

	if (mp.find(n) != mp.end())
	return mp[n];

	int res = 1;

	for (int l = 2, r; l <= n; l = r + 1) {
	r = n / (n / l);
	res = dec(res, 1ll * (r - l + 1) * Sum(n / l) % Mod);
	}

	return mp[n] = res;
	}

	signed main() {
	sieve();
	scanf("%d", &n);
	int ans = 0;

	for (int l = 1, r; l <= n; l = r + 1) {
	r = n / (n / l);
	ans = add(ans, 1ll * (n / l) * (n / l) % Mod * (n / l) % Mod * dec(Sum(r), Sum(l - 1)) % Mod);
	}

	printf("%d", ans);
	return 0;
	}

	#include <bits/stdc++.h>
	using namespace std;
	const int Mod = 1e9 + 7;
	const int N = 5e6 + 7;

	int pri[N], low[N], g[N], sum[N];
	bool isp[N];

	int T, n, m, k, pcnt;

	inline int add(int x, int y) {
	x += y;

	if (x >= Mod)
	x -= Mod;

	return x;
	}

	inline int dec(int x, int y) {
	x -= y;

	if (x < 0)
	x += Mod;

	return x;
	}

	inline int mi(int a, int b) {
	int res = 1;

	for (; b; b >>= 1, a = 1ll * a * a % Mod)
	if (b & 1)
	res = 1ll * res * a % Mod;

	return res;
	}

	inline void sieve() {
	memset(isp, true, sizeof(isp));
	isp[1] = false, low[1] = 1, g[1] = 1;

	for (int i = 2; i < N; ++i) {
	if (isp[i]) {
	pri[++pcnt] = i, low[i] = i, g[i] = dec(mi(i, k), 1);
	int x = i;

	while (1ll * x * i < N)
	x *= i, low[x] = x, g[x] = dec(mi(x, k), mi(x / i, k));
	}

	for (int j = 1; j <= pcnt && i * pri[j] < N; ++j) {
	int x = i * pri[j];
	isp[x] = false;
	low[x] = i % pri[j] ? pri[j] : low[i] * pri[j];
	g[x] = 1ll * g[x / low[x]] * g[low[x]] % Mod;

	if (!(i % pri[j]))
	break;
	}
	}

	for (int i = 1; i < N; ++i)
	sum[i] = add(sum[i - 1], g[i]);
	}

	signed main() {
	scanf("%d%d", &T, &k);
	sieve();

	while (T--) {
	scanf("%d%d", &n, &m);
	int ans = 0;

	if (n > m)
	swap(n, m);

	for (int l = 1, r; l <= n; l = r + 1) {
	r = min(n / (n / l), m / (m / l));
	ans = add(ans, 1ll * (n / l) * (m / l) % Mod * dec(sum[r], sum[l - 1]) % Mod);
	}

	printf("%d\n", ans);
	}

	return 0;
	}