高中生都能看懂的莫比乌斯反演

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)，( ext{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{ egin{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} ight.]

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) eq0)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 imes(-1)^i=sum_{i=0}^{s}C_s^i imes(-1)^i imes 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)=sumlimits_{d|n}g(d) ]

则有

[g(n)=sumlimits_{d|n}mu(d)f(frac{n}{d}) ]

证明

• 法一：对原式做数论变换

  1. (sumlimits_{d|n}g(d))替换(f(n))，即

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

  2. 变换求和顺序得

     [sumlimits_{k|n}g(k)sumlimits_{d|frac n k}mu(frac nk) ]

  3. 因为(sumlimits_{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

     [sumlimits_{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]Leftrightarrowsumlimits_{d|gcd(i,j)}mu(d))

证明

• 法一：

  设(n=gcd(i,j))，那么等式右边(=sumlimits_{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))=sumlimits_{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( ext{Id}=varphi*1)。

题

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(sumlimits_{x=a}^{b}sumlimits_{y=c}^{d}[gcd(x,y)=k])

设(f(n,m)=sumlimits_{i=1}^{n}sumlimits_{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$$sumlimits_{i = 1}^{lfloorfrac{n}{k} floor}sumlimits_{j = 1}^{lfloorfrac{m}{k} floor}[gcd(i,j) = 1]$$

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

   [sumlimits_{i=1}^{lfloorfrac{n}{k} floor}sumlimits_{j=1}^{lfloorfrac{m}{k} floor}sumlimits_{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

   [sumlimits_{d=1}^{min(n,m)}mu(d)sumlimits_{i=1}^{lfloorfrac{n}{k} floor}[d|i]sumlimits_{j=1}^{lfloorfrac{m}{k} floor}[d|j] ]

   也就是说当(d|i)且(d|j)时，(d|gcd(i,j))

4. 易得(1sim lfloorfrac{n}{k} floor)中一共有(lfloorfrac{n}{dk} floor)个(d)的倍数，同理(1sim lfloorfrac{m}{k} floor)中一共有(lfloorfrac{m}{dk} floor)个(d)的倍数，于是原式化为$$sumlimits_{d=1}^{min(n,m)}mu(d)lfloorfrac{n}{dk} floorlfloorfrac{m}{dk} floor$$

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+Tsqrt 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) << '
';
}

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

[sumlimits_{x=1}^{a}sumlimits_{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

[egin{align*}&sumlimits_{i=1}^{n}sumlimits_{j=1}^{m}[gcd(i,j)=k]\=& sumlimits_{i=1}^{lfloor frac nk floor}sumlimits_{j=1}^{lfloorfrac mk floor}[gcd(i,j)=1]\=&sumlimits_{i=1}^{lfloor frac nk floor}sumlimits_{j=1}^{lfloorfrac mk floor}sumlimits_{d|gcd(i,j)}mu(d)\=&sumlimits_{i=1}^{lfloor frac nk floor}sumlimits_{j=1}^{lfloorfrac mk floor}sumlimits_{d|i}sum _{d|j}mu(d)\=&sumlimits_{d=1}^{min(lfloorfrac nk floor,lfloorfrac mk floor)}mu(d)sumlimits_{i=1}^{lfloor frac nk floor}sumlimits_{j=1}^{lfloorfrac mk floor}sumlimits_{d|i}sumlimits_{d|j}1\=&sumlimits_{d=1}^{min(lfloorfrac nk floor,lfloorfrac mk floor)}mu(d)sumlimits_{i=1}^{lfloor frac nk floor}sumlimits_{d|i}1sumlimits_{j=1}^{lfloorfrac mk floor}sumlimits_{d|j}1\=&sumlimits_{d=1}^{min(lfloorfrac nk floor,lfloorfrac mk floor)}mu(d)lfloorfrac{lfloor frac nk floor}d floorlfloorlfloorfrac{frac mk floor}d floor\=&sumlimits_{d=1}^{min(lfloorfrac nk floor,lfloorfrac mk floor)}mu(d)lfloorfrac n{kd} floorlfloorfrac m{kd} floorend{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) << '
';
	}
	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

求

[sumlimits_{i=1}^{n}sumlimits_{j=1}^{m} ext{lcm}(i,j)(mod 20101009) ]

容易想到原式等价于

[sumlimits_{i=1}^{n}sumlimits_{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

[sumlimits_{i=1}^{n}sumlimits_{j=1}^msumlimits_{d|i,d|j,gcd(frac id,frac jd)=1}frac{i*j}d ]

变换求和顺序

[sumlimits_{d=1}^{n}dsumlimits_{i=1}^{lfloorfrac{n}{d} floor}sumlimits_{j=1}^{lfloorfrac{m}{d} floor}[gcd(i,j)=1]i*j ]

记(sum(n,m)=sumlimits_{i=1}^{n}sumlimits_{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])

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

转化为首先枚举约数

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

设(i=i'*d,j=j'*d)，则可以进一步转化

[sumlimits_{d=1}^{min(n,m)}mu(d)*d^2*sumlimits_{i=1}^{lfloorfrac{n}{d} floor}sumlimits_{j=1}^{lfloorfrac{m}{d} floor}i*j ]

前半段可以处理前缀和，后半段可以(O(1))求，设

[Q(n,m)=sumlimits_{i=1}^{n}sumlimits_{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)=sumlimits_{d=1}^{min(n,m)}mu(d)*d^2*Q(lfloorfrac nd floor,lfloorfrac md floor) ]

可以用数论分块求解

回带到原式中

[sumlimits_{d=1}^{min(n, m)}d*sum(lfloorfrac nd floor,lfloorfrac md floor) ]

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) << '
';
}

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(1leq xleq n,1leq yleq m)，且(gcd(x,y))是素数。

(n,mleq 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 << '
';
    }
    return 0;
}

洛谷 P3327 [SDOI2015]约数个数和

题目链接

求

[sumlimits_{i=1}^{n}sumlimits_{j=1}^{m}d(ij) ]

(d(x))为(x)的约数个数和

需要用到

[d(ij)=sumlimits_{x|i}sumlimits_{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 << '
';
	}
	return 0;
}

洛谷 P4449 于神之怒加强版

求

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

还是直接淦式子

[egin{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^ksum_{i=1}^{lfloorfrac nd floor}sum_{j=1}^{lfloorfrac md floor}[gcd(i,j)=1]\=&sum_{d=1}^{min(n,m)}d^ksum_{i=1}^{lfloorfrac nd floor}sum_{j=1}^{lfloorfrac md floor}sum_{x|gcd(i,j)}mu(x)\=&sum_{d=1}^{min(n,m)}d^ksum_{i=1}^{lfloorfrac nd floor}sum_{j=1}^{lfloorfrac md floor}sum_{x|i,x|j}mu(x)\=&sum_{d=1}^{min(n,m)}d^ksum_{x=1}^{min(lfloorfrac nd floor,lfloorfrac md floor)}mu(x)sum_{i=1}^{lfloorfrac nd floor}[x|i]sum_{j=1}^{lfloorfrac md floor}[x|j]\=&sum_{d=1}^{min(n,m)}d^ksum_{x=1}^{min(lfloorfrac nd floor,lfloorfrac md floor)}mu(x)lfloorfrac n{dx} floorlfloorfrac m{dx} floorend{align*} ]

令(P=dx)，则原式等于

[sum_{P=1}^{min(n,m)}lfloorfrac n{P} floorlfloorfrac m{P} floorsum_{d|P}d^kmu(frac Pd) ]

显然前面的(lfloorfrac n{P} floorlfloorfrac m{P} floor)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^kmu(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 << '
';
	}
	return 0;
}