高中生都能看懂的莫比烏斯反演

寫在前面

額，標題好像少了一個不字……應該是高中生都不能看懂的莫比烏斯反演

這是蒟蒻第一次寫這么長的博文

\(gyh\ nb\)，\(\text{OI-Wiki}\ nb\)

如果覺得寫得湊合就點個支持吧\(qwq\)

前置知識

積性函式、狄利克雷卷積、數論分塊（這一篇去找gyh吧我講也講不好）(有空慢慢補)

Mobius函式

定義

莫比烏斯函式\(\mu(n)\)定義為：

\[\mu(n)=\left\{ \begin{aligned} 1, & n=1 \\ (-1)^{s}, & n=p_1p_2…p_s（s為n的本質不同的質因子個數）\\0,&n有平方因子\end{aligned} \right.\]

其中\(p_1p_2…，p_s\)是不同素數，

可以看出，當\(n\)沒有平方因子時，\(\mu(n)\)非零，

\(\mu\)也是積性函式，

性質

莫比烏斯函式具有如下的性質：

\[\sum_{d|n}\mu(d)=\epsilon(n)=[n=1] \]

使用狄利克雷卷積來表示，即

\[\mu*1=\epsilon \]

證明：

\(n=1\)時顯然成立，

若\(n>1\)，設\(n\)有\(s\)個不同的素因子，由于\(\mu(d)\neq0\)當且僅當\(d\)無平方因子，故\(d\)中每個素因子的指數只能為\(0\)或\(1\)，

若\(n\)的某個因子\(d\)，有\(\mu(d)=(-1)^i\)，則它由\(i\)個本質不同的質因子構成，因為質因子的總數為\(s\)，所以滿足上式的因子數有\(C_s^i\)個，

所以我們就可以對于原式，轉化為列舉\(\mu(d)\)的值，同時運用二項式定理，故有

\[\sum_{d|n}\mu(d)=\sum_{i=0}^{s}C_s^i\times(-1)^i=\sum_{i=0}^{s}C_s^i\times(-1)^i\times 1^{s-i}=(1+(-1))^s \]

該式在\(s=0\)即\(n=1\)時為1，否則為\(0\)，

求莫比烏斯函式

因為莫比烏斯函式是積性函式，所以可以用線性篩

int n, cnt, p[A], mu[A];
bool vis[A];

void getmu() {
    mu[1] = 1;
    for (int i = 2; i <= n; i++) {
	if (!vis[i]) mu[i] = -1, p[++cnt] = i;
	for (int j = 1; j <= cnt && i * p[j] <= n; j++) {
	    vis[i * p[j]] = 1;
	    if (i % p[j] == 0) { mu[i * p[j]] = 0; break; }
	    mu[i * p[j]] -= mu[i];
	}
    }
}

莫比烏斯反演公式

設\(f(n)\)，\(g(n)\)為兩個數論函式，

如果有

\[f(n)=\sum\limits_{d|n}g(d) \]

則有

\[g(n)=\sum\limits_{d|n}\mu(d)f(\frac{n}{d}) \]

證明

• 法一：對原式做數論變換

  1. \(\sum\limits_{d|n}g(d)\)替換\(f(n)\)，即

     \[\sum\limits_{d|n}\mu(d)f(\frac{n}{d})=\sum\limits_{d|n}\mu(d)\sum\limits_{k|\frac nd}g(k) \]

  2. 變換求和順序得

     \[\sum\limits_{k|n}g(k)\sum\limits_{d|\frac n k}\mu(\frac nk) \]

  3. 因為\(\sum\limits_{d|n}\mu(d)=[n=1]\)，所以只有在\(\frac{n}{k}=1\)即\(n=k\)時才會成立，所以上式等價于

     \[\sum\limits_{d|n}[n=k]g(k)=g(n) \]

  得證

• 法二：利用卷積

  原問題為：已知\(f=g*1\)，證明\(g=f*\mu\)

  轉化：\(f*\mu=g*1*\mu=g*\varepsilon=g\)（\(1*\mu=\varepsilon\)）

  再次得證= =

小性質

\([\gcd(i,j)=1]\Leftrightarrow\sum\limits_{d|\gcd(i,j)}\mu(d)\)

證明

• 法一：

  設\(n=\gcd(i,j)\)，那么等式右邊\(=\sum\limits_{d|n}\mu(d)=[n=1]=[\gcd(i,j)=1]=\)等式左邊

• 法二：

  利用單位函式暴力拆開：\([\gcd(i,j)=1]=\varepsilon(\gcd(i,j))=\sum\limits_{d|\gcd(i,j)}\mu(d)\)

做題思路&&應用

利用狄利克雷卷積可以解決一系列求和問題，常見做法是使用一個狄利克雷卷積替換求和式中的一部分，然后調換求和順序，最終降低時間復雜度，

經常利用的卷積有\(\mu*1=\epsilon\)和\(\text{Id}=\varphi*1\)，

題

還是以題為主吧，但是做的題也會單獨寫題解，畢竟要多水幾篇博客的嘛/huaji

洛谷 P2522 [HAOI2011]Problem b

題目鏈接

我的題解

有\(n\)組詢問，每次給出\(a,b,c,d,k\)，求\(\sum\limits_{x=a}^{b}\sum\limits_{y=c}^{d}[\gcd(x,y)=k]\)

設\(f(n,m)=\sum\limits_{i=1}^{n}\sum\limits_{j= 1}^{m}[\gcd(i,j)=k]\)

那么根據容斥原理，題目中的式子就轉化成了\(f(b,d)-f(b, c - 1) - f(a - 1,d) + f(a - 1, c - 1)\)

所以我們接下來的問題就轉化為了如何求\(f\)的值

現在來化簡\(f\)的值

1. 容易得出原式等價于$$\sum\limits_{i = 1}^{\lfloor\frac{n}{k}\rfloor}\sum\limits_{j = 1}^{\lfloor\frac{m}{k}\rfloor}[\gcd(i,j) = 1]$$

2. 因為\(\epsilon(n) =\sum\limits_{d|n}\mu(d)=[n=1]\)，由此可將原式化為

   \[\sum\limits_{i=1}^{\lfloor\frac{n}{k}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{k}\rfloor}\sum\limits_{d|gcd(i,j)}\mu(d) \]

3. 改變列舉物件并改變列舉順序，先列舉\(d\)，得

   \[\sum\limits_{d=1}^{\min(n,m)}\mu(d)\sum\limits_{i=1}^{\lfloor\frac{n}{k}\rfloor}[d|i]\sum\limits_{j=1}^{\lfloor\frac{m}{k}\rfloor}[d|j] \]

   也就是說當\(d|i\)且\(d|j\)時，\(d|\gcd(i,j)\)

4. 易得\(1\sim \lfloor\frac{n}{k}\rfloor\)中一共有\(\lfloor\frac{n}{dk}\rfloor\)個\(d\)的倍數，同理\(1\sim \lfloor\frac{m}{k}\rfloor\)中一共有\(\lfloor\frac{m}{dk}\rfloor\)個\(d\)的倍數，于是原式化為$$\sum\limits_{d=1}^{\min(n,m)}\mu(d)\lfloor\frac{n}{dk}\rfloor\lfloor\frac{m}{dk}\rfloor$$

此時已經可以\(O(n)\)求解，但是過不了，因為有很多值相同的區間，所以可以用數論分塊來做

先預處理\(\mu\)，再用數論分塊，復雜度\(O(n+T\sqrt n)\)

我的代碼每次得分玄學，看評測機心情，建議自己寫

/*
Author:loceaner
*/
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int A = 1e6 + 11;
const int B = 1e6 + 11;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;

inline int read() {
	char c = getchar(); int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
	return x * f;
}

int n, a, b, c, d, k, cnt, p[A], mu[A], sum[A];
bool vis[A];

void getmu() {
	int MAX = 50010;
	mu[1] = 1;
	for (int i = 2; i <= MAX; i++) {
		if (!vis[i]) mu[i] = -1, p[++cnt] = i;
		for (int j = 1; j <= cnt && i * p[j] <= MAX; j++) {
			vis[i * p[j]] = true;
			if (i % p[j] == 0) break;
			mu[i * p[j]] -= mu[i];
		}
	}
	for (int i = 1; i <= MAX; i++) sum[i] = sum[i - 1] + mu[i];
}

int work(int x, int y) {
	int ans = 0ll;
	int max = min(x, y);
	for (int l = 1, r; l <= max; l = r + 1) {
		r = min(x / (x / l), y / (y / l));
		ans += (1ll * x / (l * k)) * (1ll * y / (l * k)) * 1ll * (sum[r] - sum[l - 1]); 
	}
	return ans;
}

void solve() {
	a = read(), b = read(), c = read(), d = read(), k = read();
	cout << work(b, d) - work(a - 1, d) - work(b, c - 1) + work(a - 1, c - 1) << '\n';
}

signed main() {
	getmu();
	int T = read();
	while (T--) solve();
	return 0;
}

洛谷 P3455 [POI2007]ZAP-Queries

題目鏈接

我的題解

有\(T\)組詢問，每次詢問求

\[\sum\limits_{x=1}^{a}\sum\limits_{y=1}^{b}[\gcd(x,y)=d] \]

因為我不喜歡用\(x、y、a、b、d\)，所以一一對應換成\(i、j、n、m、k\)

直接淦式子

\[\begin{align*}&\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}[\gcd(i,j)=k]\\=& \sum\limits_{i=1}^{\lfloor \frac nk\rfloor}\sum\limits_{j=1}^{\lfloor\frac mk\rfloor}[\gcd(i,j)=1]\\=&\sum\limits_{i=1}^{\lfloor \frac nk\rfloor}\sum\limits_{j=1}^{\lfloor\frac mk\rfloor}\sum\limits_{d|\gcd(i,j)}\mu(d)\\=&\sum\limits_{i=1}^{\lfloor \frac nk\rfloor}\sum\limits_{j=1}^{\lfloor\frac mk\rfloor}\sum\limits_{d|i}\sum _{d|j}\mu(d)\\=&\sum\limits_{d=1}^{\min(\lfloor\frac nk\rfloor,\lfloor\frac mk\rfloor)}\mu(d)\sum\limits_{i=1}^{\lfloor \frac nk\rfloor}\sum\limits_{j=1}^{\lfloor\frac mk\rfloor}\sum\limits_{d|i}\sum\limits_{d|j}1\\=&\sum\limits_{d=1}^{\min(\lfloor\frac nk\rfloor,\lfloor\frac mk\rfloor)}\mu(d)\sum\limits_{i=1}^{\lfloor \frac nk\rfloor}\sum\limits_{d|i}1\sum\limits_{j=1}^{\lfloor\frac mk\rfloor}\sum\limits_{d|j}1\\=&\sum\limits_{d=1}^{\min(\lfloor\frac nk\rfloor,\lfloor\frac mk\rfloor)}\mu(d)\lfloor\frac{\lfloor \frac nk\rfloor}d\rfloor\lfloor\lfloor\frac{\frac mk\rfloor}d\rfloor\\=&\sum\limits_{d=1}^{\min(\lfloor\frac nk\rfloor,\lfloor\frac mk\rfloor)}\mu(d)\lfloor\frac n{kd}\rfloor\lfloor\frac m{kd}\rfloor\end{align*} \]

現在就可以每次詢問\(O(n)\)做這道題了

但是跑不過啊，不過顯然可以數論分塊，所以我們就可以\(O(\sqrt n)\)回答每次詢問了

/*
Author:loceaner
*/
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int A = 5e5 + 11;
const int B = 1e6 + 11;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;

inline int read() {
	char c = getchar(); int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
	return x * f;
}

int n, m, k, mu[A], p[A], sum[A], cnt;
bool vis[A];

void getmu(int n) {
	mu[1] = 1;
	for (int i = 2; i <= n; i++) {
		if (!vis[i]) p[++cnt] = i, mu[i] = -1;
		for (int j = 1; j <= cnt && i * p[j] <= n; j++) {
			vis[i * p[j]] = 1;
			if (i % p[j] == 0) break;
			mu[i * p[j]] -= mu[i];
		}
	}
	for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + mu[i];
}

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

int main() {
	getmu(50000);
	int T = read();
	while (T--) {
		n = read(), m = read(), k = read();
		cout << solve(n, m, k) << '\n';
	}
	return 0;
}

洛谷 P1829 [國家集訓隊]Crash的數字表格 / JZPTAB

題目鏈接

我的題解

求

\[\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}\text{lcm}(i,j)(\bmod 20101009) \]

容易想到原式等價于

\[\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}\frac{ij}{\gcd(i,j)} \]

列舉 \(i,j\) 的最大公約數 \(d\)，顯然 \(\gcd(\frac id,\frac jd)=1\)，即 \(\frac id\) 和 \(\frac jd\) 互質

\[\sum\limits_{i=1}^{n}\sum\limits_{j=1}^m\sum\limits_d[d|i][d|j][\gcd(\frac id,\frac jd)=1]\frac{ij}d \]

變換求和順序，改為先列舉 \(d\)，則可以得到：

\[\sum\limits_{d=1}\sum\limits_{i=1}^{n}[d|i]\sum\limits_{j=1}^{m}[d|j][\gcd(\frac id,\frac jd)=1]\frac{ij}{d} \]

把后面的 \(d\) 提取出來，對應的兩個 \(\sum\) 改為列舉 \(\frac id,\frac jd\)，進一步化簡可以得到：

\[\begin{aligned}&\sum\limits_{d=1}\sum\limits_{i=1}^{n}[d|i]\sum\limits_{j=1}^{m}[d|j][\gcd(\frac id,\frac jd)=1]\frac{ij}{d}\\=&\sum\limits_{d=1}\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{d}\rfloor}[\gcd(i,j)=1]\frac{id\times jd}{d}\\=&\sum\limits_{d=1}\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{d}\rfloor}[\gcd(i,j)=1]ijd\\=&\sum\limits_{d=1}^{n}d\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{d}\rfloor}[\gcd(i,j)=1]ij\end{aligned} \]

先擱置這個式子，考慮后半部分，

記 \(sum(n,m)=\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}[\gcd(i,j)=1]ij\)，對其進行化簡，用 \(\varepsilon(\gcd(i,j))\)（即\(\sum\limits_{d|、gcd(i,j)}\mu(d)\)） 替換\([\gcd(i,j)=1]\)，得到

\[\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}\sum\limits_{d|\gcd(i,j)}\mu(d)ij \]

轉化為首先列舉約數

\[\sum\limits_{d=1}^{\min(n,m)}\mu(d)\sum\limits_{i=1}^{n}[d|i]\sum\limits_{j=1}^{m}[d|j]ij \]

像上面一樣，后面的式子中考慮把 \(d\) 提取出來，直接列舉滿足條件的數，得到

\[\begin{aligned}&\sum\limits_{d=1}^{\min(n,m)}\mu(d)\sum\limits_{i=1}^{n}[d|i]\sum\limits_{j=1}^{m}[d|j]ij\\&=\sum\limits_{d=1}^{\min(n,m)}\mu(d)\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{d}\rfloor}id\times jd\\&=\sum\limits_{d=1}^{\min(n,m)}\mu(d)\times d^2\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{d}\rfloor}ij\end{aligned} \]

前半段可以處理前綴和，后半段可以 \(O(1)\) 求，因為設

\[Q(n,m)=\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}ij=\frac{n\times(n+1)}{2}\times\frac{m\times(m+1)}{2} \]

的話，這個函式顯然可以\(O(1)\)求解

到現在

\[sum(n,m)=\sum\limits_{d=1}^{\min(n,m)}\mu(d)\times d^2\times Q(\lfloor\frac nd \rfloor,\lfloor\frac md\rfloor) \]

可以用數論分塊求解

回帶到原式中

\[\sum\limits_{d=1}^{\min(n, m)}d\times sum(\lfloor\frac nd \rfloor,\lfloor\frac md\rfloor) \]

這樣的話，又可以數論分塊求解了，然后就做完啦

/*
Author:loceaner
*/
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define int long long
using namespace std;

const int A = 1e7 + 11;
const int B = 1e6 + 11;
const int mod = 20101009;
const int inf = 0x3f3f3f3f;

inline int read() {
	char c = getchar(); int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
	return x * f;
}

bool vis[A];
int n, m, mu[A], p[B], sum[A], cnt;

void getmu() {
    mu[1] = 1;
    int k = min(n, m);
    for (int i = 2; i <= k; i++) {
        if (!vis[i]) p[++cnt] = i, mu[i] = -1;
        for (int j = 1; j <= cnt && i * p[j] <= k; ++j) {
            vis[i * p[j]] = 1;
            if (i % p[j] == 0) break;
            mu[i * p[j]] = -mu[i];
        }
    }
    for (int i = 1; i <= k; i++) sum[i] = (sum[i - 1] + i * i % mod * mu[i]) % mod;
}

int Sum(int x, int y) { 
	return (x * (x + 1) / 2 % mod) * (y * (y + 1) / 2 % mod) % mod; 
}

int solve2(int x, int y) {
    int res = 0;
    for (int i = 1, j; i <= min(x, y); i = j + 1) {
        j = min(x / (x / i), y / (y / i));
        res = (res + 1LL * (sum[j] - sum[i - 1] + mod) * Sum(x / i, y / i) % mod) % mod;
    }
    return res;
}

int solve(int x, int y) {
    int res = 0;
    for (int i = 1, j; i <= min(x, y); i = j + 1) {
        j = min(x / (x / i), y / (y / i));
        res = (res + 1LL * (j - i + 1) * (i + j) / 2 % mod * solve2(x / i, y / i) % mod) % mod;
    }
    return res;
}

signed main() {
    n = read(), m = read();
    getmu();
    cout << solve(n, m) << '\n';
}

洛谷 P2257 YY的GCD

題目鏈接

給定\(n,m\),求二元組\((x,y)\)的個數，滿足\(1\leq x\leq n,1\leq y\leq m\)，且\(gcd(x,y)\)是素數，

\(n,m\leq 10^7\)，自帶多組資料，至多\(10^{4}\)組資料，

思路與第一題Problem B類似，在這里不再贅述，只給出代碼= =

#include <cmath> 
#include <cstdio>
#include <cstring> 
#include <iostream>
using namespace std;

const int A = 1e7 + 11; 

inline int read() {
	char c = getchar(); int x = 0, f = 1;
	for ( ; !isdigit(c); c = getchar()) if(c == '-') f = -1; 
	for ( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
	return x * f;
}

bool vis[A];
long long sum[A];
int prim[A], mu[A], g[A], cnt, n, m;

void get_mu(int n) {
    mu[1] = 1;
    for (int i = 2; i <= n; i++) {
        if (!vis[i]) {
            mu[i] = -1;
            prim[++cnt] = i;
        }
        for (int j = 1; j <= cnt && prim[j] * i <= n; j++) {
            vis[i * prim[j]] = 1;
            if (i % prim[j] == 0) break;
            else mu[prim[j] * i] = - mu[i];
        }
    }
    for (int j = 1; j <= cnt; j++)
        for (int i = 1; i * prim[j] <= n; i++) g[i * prim[j]] += mu[i];
    for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + (long long)g[i];
}

signed main() {
    int t = read();
    get_mu(10000000);
    while (t--) {
        n = read(), m = read();
        if (n > m) swap(n, m);
        long long ans = 0;
        for (int l = 1, r; l <= n; l = r + 1) {
            r = min(n / (n / l), m / (m / l));
            ans += 1ll * (n / l) * (m / l) * (sum[r] - sum[l - 1]);
        }
        cout << ans << '\n';
    }
    return 0;
}

洛谷 P3327 [SDOI2015]約數個數和

題目鏈接

求

\[\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}d(ij) \]

\(d(x)\)為\(x\)的約數個數和

需要用到

\[d(ij)=\sum\limits_{x|i}\sum\limits_{y|j}[\gcd(x,y)=1] \]

證明我也不會

然后自己推導吧，在此不再贅述

/*
Author:loceaner
*/
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int A = 5e5 + 11;
const int B = 1e6 + 11;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;

inline int read() {
	char c = getchar(); int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
	return x * f;
}

bool vis[A];
int n, m, p[A], mu[A], cnt, sum[A];
long long g[A], ans;

void getmu(int n) {
	mu[1] = 1;
	for (int i = 2; i <= n; i++) {
		if (!vis[i]) mu[i] = -1, p[++cnt] = i;
		for (int j = 1; j <= cnt && i * p[j] <= n; j++) {
			vis[i * p[j]] = 1;
			if (i % p[j] == 0) break;
			mu[i * p[j]] -= mu[i];
		}
	}
	for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + mu[i];
	for (int i = 1; i <= n; i++) {
		int ans = 0;
		for (int l = 1, r; l <= i; l = r + 1) {
			r = (i / (i / l));
			ans += 1ll * (r - l + 1) * (i / l);
		}
		g[i] = ans;
	}
}

signed main() {
	int T = read();
	getmu(50000);
	while (T--) {
		n = read(), m = read();
		int maxn = min(n, m);
		ans = 0;
		for (int l = 1, r; l <= maxn; l = r + 1) {
			r = min(n / (n / l), m / (m / l));
			ans += 1ll * (sum[r] - sum[l - 1]) * 1ll * g[n / l] * 1ll * g[m / l];
		}
		cout << ans << '\n';
	}
	return 0;
}

洛谷 P4449 于神之怒加強版

求

\[\sum_{i=1}^{n}\sum_{j=1}^{m}\gcd(i,j)^k(\bmod 1e9+7) \]

還是直接淦式子

\[\begin{align*}&\sum_{i=1}^{n}\sum_{j=1}^{m}\gcd(i,j)^k\\=&\sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{d=1}^{\min(n,m)}d^k[\gcd(i,j)=d]\\=&\sum_{d=1}^{\min(n,m)}d^k\sum_{i=1}^{\lfloor\frac nd\rfloor}\sum_{j=1}^{\lfloor\frac md\rfloor}[\gcd(i,j)=1]\\=&\sum_{d=1}^{\min(n,m)}d^k\sum_{i=1}^{\lfloor\frac nd\rfloor}\sum_{j=1}^{\lfloor\frac md\rfloor}\sum_{x|\gcd(i,j)}\mu(x)\\=&\sum_{d=1}^{\min(n,m)}d^k\sum_{i=1}^{\lfloor\frac nd\rfloor}\sum_{j=1}^{\lfloor\frac md\rfloor}\sum_{x|i,x|j}\mu(x)\\=&\sum_{d=1}^{\min(n,m)}d^k\sum_{x=1}^{\min(\lfloor\frac nd\rfloor,\lfloor\frac md\rfloor)}\mu(x)\sum_{i=1}^{\lfloor\frac nd\rfloor}[x|i]\sum_{j=1}^{\lfloor\frac md\rfloor}[x|j]\\=&\sum_{d=1}^{\min(n,m)}d^k\sum_{x=1}^{\min(\lfloor\frac nd\rfloor,\lfloor\frac md\rfloor)}\mu(x)\lfloor\frac n{dx}\rfloor\lfloor\frac m{dx}\rfloor\end{align*} \]

令\(P=dx\)，則原式等于

\[\sum_{P=1}^{\min(n,m)}\lfloor\frac n{P}\rfloor\lfloor\frac m{P}\rfloor\sum_{d|P}d^k\mu(\frac Pd) \]

顯然前面的\(\lfloor\frac n{P}\rfloor\lfloor\frac m{P}\rfloor\)部分可以分塊求解，

現在考慮后面的一部分，令

\[g(n)=\sum_{d|n}d^k\mu(\frac nd) \]

容易得出這個函式是積性函式，所以我們就可以線性篩然后求出其前綴和

然后就做完了

/*
Author:loceaner
莫比烏斯反演
*/
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define int long long
using namespace std;

const int A = 5e6 + 11;
const int B = 1e6 + 11;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;

inline int read() {
	char c = getchar();
	int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
	return x * f;
}

bool vis[A];
int T, n, m, k, f[A], g[A], p[A], cnt, sum[A];

inline int power(int a, int b) {
	int res = 1;
	while (b) {
		if (b & 1) res = res * a % mod;
		a = a * a % mod, b >>= 1;
	}
	return res;
}

inline int mo(int x) {
	if(x > mod) x -= mod;
	return x;
}

inline void work() {
	g[1] = 1;
	int maxn = 5e6 + 1;
	for (int i = 2; i <= maxn; i++) {
		if (!vis[i]) { p[++cnt] = i, f[cnt] = power(i, k), g[i] = mo(f[cnt] - 1 + mod); }
		for (int j = 1; j <= cnt && i * p[j] <= maxn; j++) {
			vis[i * p[j]] = 1;
			if (i % p[j] == 0) { g[i * p[j]] = g[i] * 1ll * f[j] % mod; break; }
			g[i * p[j]] = g[i] * 1ll * g[p[j]] % mod;
		}
	}
	for (int i = 2; i <= maxn; i++) g[i] = (g[i - 1] + g[i]) % mod;
}

inline int abss(int x) {
	while (x < 0) x += mod;
	return x;
}

signed main() {
	T = read(), k = read();
	work();
	while (T--) {
		n = read(), m = read();
		int maxn = min(n, m), ans = 0;
		for (int l = 1, r; l <= maxn; l = r + 1) {
			r = min(n / (n / l), m / (m / l));
			(ans += abss(g[r] - g[l - 1]) * 1ll * (n / l) % mod * (m / l) % mod) %= mod;
		}
		ans = (ans % mod + mod) % mod;
		cout << ans << '\n';
	}
	return 0;
}

標籤：C++

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