高中生都能看懂的莫比乌斯反演

0
10

写在前面

额,标题好像少了一个不字……应该是高中生都不能看懂的莫比乌斯反演

这是蒟蒻第一次写这么长的博文

\(gyh\ nb\),\(\text{OI-Wiki}\ nb\)

如果觉得写得凑合就点个支持吧\(qwq\)

前置知识

积性函数、狄利克雷卷积、数论分块(这一篇去找gyh吧我讲也讲不好)(有空慢慢补)

Mobius函数

定义

莫比乌斯函数\(\mu(n)\)定义为:

\[\mu(n)=\left\{
\begin{aligned}
1, & n=1 \\
(-1)^{s}, & n=p_1p_2…p_s(s为n的本质不同的质因子个数)\\0,&n有平方因子\end{aligned}
\right.\]

其中\(p_1p_2…,p_s\)是不同素数。

可以看出,当\(n\)没有平方因子时,\(\mu(n)\)非零。

\(\mu\)也是积性函数。

性质

莫比乌斯函数具有如下的性质:

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

使用狄利克雷卷积来表示,即

\[\mu*1=\epsilon
\]

证明:

\(n=1\)时显然成立。

若\(n>1\),设\(n\)有\(s\)个不同的素因子,由于\(\mu(d)\neq0\)当且仅当\(d\)无平方因子,故\(d\)中每个素因子的指数只能为\(0\)或\(1\)。

若\(n\)的某个因子\(d\),有\(\mu(d)=(-1)^i\),则它由\(i\)个本质不同的质因子构成。因为质因子的总数为\(s\),所以满足上式的因子数有\(C_s^i\)个。

所以我们就可以对于原式,转化为枚举\(\mu(d)\)的值,同时运用二项式定理,故有

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

该式在\(s=0\)即\(n=1\)时为1,否则为\(0\)。

求莫比乌斯函数

因为莫比乌斯函数是积性函数,所以可以用线性筛

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];
	}
    }
}

莫比乌斯反演公式

设\(f(n)\),\(g(n)\)为两个数论函数。

如果有

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

则有

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

证明

  • 法一:对原式做数论变换

    1. \(\sum\limits_{d|n}g(d)\)替换\(f(n)\),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. 变换求和顺序得

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

    3. 因为\(\sum\limits_{d|n}\mu(d)=[n=1]\),所以只有在\(\frac{n}{k}=1\)即\(n=k\)时才会成立,所以上式等价于

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

    得证

  • 法二:利用卷积

    原问题为:已知\(f=g*1\),证明\(g=f*\mu\)

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

    再次得证= =

小性质

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

证明

  • 法一:

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

  • 法二:

    利用单位函数暴力拆开:\([\gcd(i,j)=1]=\varepsilon(\gcd(i,j))=\sum\limits_{d|\gcd(i,j)}\mu(d)\)

做题思路&&应用

利用狄利克雷卷积可以解决一系列求和问题。常见做法是使用一个狄利克雷卷积替换求和式中的一部分,然后调换求和顺序,最终降低时间复杂度。

经常利用的卷积有\(\mu*1=\epsilon\)和\(\text{Id}=\varphi*1\)。

还是以题为主吧,但是做的题也会单独写题解,毕竟要多水几篇博客的嘛/huaji

洛谷 P2522 [HAOI2011]Problem b

题目链接

我的题解

有\(n\)组询问,每次给出\(a,b,c,d,k\),求\(\sum\limits_{x=a}^{b}\sum\limits_{y=c}^{d}[\gcd(x,y)=k]\)

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

那么根据容斥原理,题目中的式子就转化成了\(f(b,d)-f(b, c – 1) – f(a – 1,d) + f(a – 1, c – 1)\)

所以我们接下来的问题就转化为了如何求\(f\)的值

现在来化简\(f\)的值

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

  2. 因为\(\epsilon(n) =\sum\limits_{d|n}\mu(d)=[n=1]\),由此可将原式化为

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

  3. 改变枚举对象并改变枚举顺序,先枚举\(d\),得

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

    也就是说当\(d|i\)且\(d|j\)时,\(d|\gcd(i,j)\)

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

此时已经可以\(O(n)\)求解,但是过不了,因为有很多值相同的区间,所以可以用数论分块来做

先预处理\(\mu\),再用数论分块,复杂度\(O(n+T\sqrt n)\)

我的代码每次得分玄学,看评测机心情,建议自己写

/*
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

题目链接

我的题解

有\(T\)组询问,每次询问求

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

因为我不喜欢用\(x、y、a、b、d\),所以一一对应换成\(i、j、n、m、k\)

直接淦式子

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

现在就可以每次询问\(O(n)\)做这道题了

但是跑不过啊,不过显然可以数论分块,所以我们就可以\(O(\sqrt n)\)回答每次询问了

/*
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 [国家集训队]Crash的数字表格 / JZPTAB

题目链接

我的题解

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

容易想到原式等价于

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

枚举\(i,j\)的最大公约数\(d\),显然\(\gcd(\frac id,\frac jd)=1\),即\(\frac id\)和\(\frac jd\)互质

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

变换求和顺序

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

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

对其进行化简,用\(\varepsilon(\gcd(i,j))\)替换\([\gcd(i,j)=1]\)

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

转化为首先枚举约数

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

设\(i=i’*d,j=j’*d\),则可以进一步转化

\[\sum\limits_{d=1}^{\min(n,m)}\mu(d)*d^2*\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{d}\rfloor}i*j
\]

前半段可以处理前缀和,后半段可以\(O(1)\)求,设

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

显然可以\(O(1)\)求解

到现在

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

可以用数论分块求解

回带到原式中

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

又可以数论分块求解了

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

题目链接

给定\(n,m\),求二元组\((x,y)\)的个数,满足\(1\leq x\leq n,1\leq y\leq m\),且\(gcd(x,y)\)是素数。

\(n,m\leq 10^7\),自带多组数据,至多\(10^{4}\)组数据。

思路与第一题Problem B类似,在这里不再赘述,只给出代码= =

#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]约数个数和

题目链接

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

\(d(x)\)为\(x\)的约数个数和

需要用到

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

证明我也不会

然后自己推导吧,在此不再赘述

/*
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 于神之怒加强版

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

还是直接淦式子

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

令\(P=dx\),则原式等于

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

显然前面的\(\lfloor\frac n{P}\rfloor\lfloor\frac m{P}\rfloor\)部分可以分块求解。

现在考虑后面的一部分,令

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

容易得出这个函数是积性函数,所以我们就可以线性筛然后求出其前缀和

然后就做完了

/*
Author:loceaner
莫比乌斯反演
*/
#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;
}

<

发布回复

请输入评论!
请输入你的名字