P3768 简单的数学题 [狄利克雷卷积,杜教筛,莫比乌斯反演]

简单的数学题

题目连接

https://www.luogu.org/problemnew/show/P3768

题目描述

输入一个正整数$n,n\le 10^{10}$和$p,p \le 1.1 \times 10^9$.且$p$为质数.

计算$\sum_{i=1}^n\sum_{j=1}^nijgcd(i,j)$对$p$取模.

题解

方法一.暴力推公式

1. 考虑把枚举$gcd(i,j)$
   原式=$\sum_{d=1}^nd^3\sum_{i=1}^{n/d}\sum_{j=1}^{n/d}ij\times [gcd(i,j)=1]$
2. 莫比乌斯反演处理后边的式子
   设$f(x)=\sum_{i=1}^{n/d}\sum_{j=1}^{n/d}ij\times [gcd(i,j)=x]$
   $g(x) = \sum_{x|d}f(d)=x^2\sum_{i=1}^{n/dx}\sum_{j=1}^{n/dx}ij$
   故$f(x)=\sum_{x|p}\mu(p/x)g(p)=\sum_{x|p}\mu(p/x)p^2\sum_{i=1}^{n/dp}\sum_{j=1}^{n/dp}ij$
   因此$f(1)=\sum_{p=1}^n\mu(p)p^2\sum_{i=1}^{n/dp}\sum_{j=1}^{n/dp}ij$
3. 合并两部分
   原式=$\sum_{d=1}^nd^3\sum_{p=1}^n\mu(p)p^2\sum_{i=1}^{n/dp}\sum_{j=1}^{n/dp}ij$
   转而枚举$a=dp$.我们得到:
   原式$=\sum_{a=1}^n\sum_{d|a}a^2d\mu(a/d)\sum_{i=1}^{n/a}i\sum_{j=1}^{n/a}j$
   $=\sum_{a=1}^n (\frac{\lfloor \frac{n}{a} \rfloor(\lfloor \frac{n}{a} \rfloor+1)}{2})^2a^2\sum_{d|a}d\mu(a/d)$
   我们发现$Id*\mu=\phi$是常见的狄利克雷卷积.
   因此原式$=\sum_{a=1}^n (\frac{\lfloor \frac{n}{a} \rfloor(\lfloor \frac{n}{a} \rfloor+1)}{2})^2a^2\phi(a)$
   化简到这一步的时候,前面部分可以分块计算,后面的部分$a^2\phi(a)$要快速计算前缀和.
   于是我们想到了杜教筛:
   记$f(x)=x^2\phi(x)$,找到积性函数$g(x)=x^2$与它做卷积使得$g(x)$的前缀和可以快速求出,而$(f*g)(x)=\sum_{d|x}\frac{x}{d}^2\times d^2\phi(d)=x^2\sum_{d|x}\phi(x)=x^3$的前缀和也可以快速求出.
   记$S(x)=\sum_{i=1}^xf(x)$
   根据杜教筛公式$g(1)S(n)=\sum_{x=1}^n(f*g)(x) - \sum_{x=2}^ng(x)S(\frac{n}{x})$.
   $S(n)=\sum_{x=1}^nx^3-\sum_{x=2}^nx^2S(\frac{n}{x})$
   写到这就完了.

方法二.$\phi$卷积

根据公式:

$gcd(i,j) = \sum_{d|gcd(i,j)}\phi(d)=\sum_{d|i,d|j}\phi(d)$

因此

$\sum_{i=1}^n\sum_{j=1}^nijgcd(i,j)=\sum_{i=1}^n\sum_{j=1}^nij\sum_{d|i,d|j}\phi(d)=\sum_{d=1}^n\phi(d)d^2\sum_{i=1}^{n/d}\sum_{j=1}^{n/d}ij=\sum_{d=1}^n\phi(d)d^2(\frac{\lfloor \frac{n}{d} \rfloor(\lfloor \frac{n}{d} \rfloor+1)}{2})^2$

同样也得到了我们的式子,是不是推导方法简单多了?

遇见$gcd(i,j)$的时候,多想想是不是可以用$\phi$卷积来做?

可能会减少很多不必要的推导过程.

代码

#include <iostream>
#include <algorithm>
#include <cstring>
#include <unordered_map>
#define pr(x) std::cout << #x << ':' << x << std::endl
#define rep(i,a,b) for(LL i = a;i <= b;++i)
const int N = 1e7;
typedef long long LL;
LL n,p; 
LL phi[N+10];
int prime[N+10],zhi[N+10],low[N+10],pcnt;
LL mod_pow(LL x,LL n) {
    LL res = 1;
    while(n) {
        if(n&1) res = res * x % p;
        x = x * x % p;
        n >>= 1;
    }
    return res;
}
LL inv6,inv2,inv4;
void sieve() {
    pcnt = 0;
    low[1] = phi[1] = zhi[1] = 1;
    rep(i,2,N) {
        if(!zhi[i]) {
            phi[i] = i-1;
            prime[pcnt++] = i;
            low[i] = i;
        }
        for(LL j = 0;j < pcnt && prime[j]*i <= N;++j) {
            zhi[i*prime[j]] = 1;
            if(i % prime[j] == 0) {
                low[i*prime[j]] = low[i] * prime[j];
                if(i == low[i]) {
                    phi[i*prime[j]] = phi[i]*prime[j];
                }
                else {
                    phi[i*prime[j]] = phi[i/low[i]] * phi[low[i]*prime[j]];
                }
                break;
            }
            else{
                low[i*prime[j]] = prime[j];
                phi[i*prime[j]] = phi[i] * phi[prime[j]];
            }
        }

    }
}
LL sum3(LL n) {
    n %= p;
    LL res = n*(n+1)%p*(2*n+1)%p*inv6%p;
    return res;
}
LL sum4(LL n) {
    n %= p;
    LL x = n*(n+1)%p;
    return x*x%p*inv4%p;
}
std::unordered_map<LL,LL> vis,rec;
LL F(LL n) {
    if(n <= N) return phi[n];
    if(vis[n]) return rec[n];
    LL res = sum4(n);
    for(LL x = 2,last;x <= n;x = last) {
        last = n/(n/x)+1;
        res = (res - ((sum3(last-1)-sum3(x-1)+p)%p*F(n/x)%p) + p) % p;
    }
    vis[n] = 1;
    return rec[n] = res;
}

signed main() {
    sieve();
    std::ios::sync_with_stdio(false);
    std::cin >> p >> n;
    inv6 = mod_pow(6,p-2);
    inv4 = mod_pow(4,p-2);
    inv2 = mod_pow(2,p-2);
    rep(i,1,N) phi[i] = ((phi[i]*i%p*i%p) + phi[i-1]) % p;
    LL last,ans = 0;
    for(LL x = 1;x <= n;x = last+1) {
        last = n/(n/x);
        LL y = n/x%p;
        LL z = (1+y)*(y)/2%p;
        ans = (ans + (z*z%p*((F(last)-F(x-1)+p)%p))) % p;
    }
    std::cout << ans << std::endl;
    return 0;
}