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${\sum }_{i=1}^n{\sum }_{j=1}^{n}ijgcd(i,j)$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=${\sum }_{d=1}^n{d}^3{\sum }_{i=1}^{n/d}{\sum }_{j=1}^{n/d}ij\times [gcd(i,j)=1]$
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=${\sum }_{d=1}^n{d}^3{\sum }_{p=1}^n\mu (p)p^2{\sum }_{i=1}^{n/dp}{\sum }_{j=1}^{n/dp}ij$
   转而枚举$a=dp$.我们得到:
   原式$={\sum }_{a=1}^n{\sum }_{d|a}{a}^{2}d\mu (a/d){\sum }_{i=1}^{n/a}i{\sum }_{j=1}^{n/a}j$
   $={\sum }_{a=1}^n(\frac{\lfloor \frac{n}{a}\rfloor (\lfloor \frac{n}{a}\rfloor +1)}{2}{)}^2{a}^2{\sum }_{d|a}d\mu (a/d)$
   我们发现$Id\ast \mu =\varphi$是常见的狄利克雷卷积.
   因此原式$={\sum }_{a=1}^n(\frac{\lfloor \frac{n}{a}\rfloor (\lfloor \frac{n}{a}\rfloor +1)}{2}{)}^2{a}^2\varphi (a)$
   化简到这一步的时候,前面部分可以分块计算,后面的部分$a^2\varphi (a)$要快速计算前缀和.
   于是我们想到了杜教筛:
   记$f(x)=x^2\varphi (x)$,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$(f\ast g)(x)={\sum }_{d|x}{\frac{x}{d}}^2\times d^2\varphi (d)=x^2{\sum }_{d|x}\varphi (x)=x^3$的前缀和也可以快速求出.
   记$S(x)={\sum }_{i=1}^{x}f(x)$
   根据杜教筛公式$g(1)S(n)={\sum }_{x=1}^n(f\ast g)(x)-{\sum }_{x=2}^{n}g(x)S(\frac{n}{x})$.
   $S(n)={\sum }_{x=1}^n{x}^3-{\sum }_{x=2}^n{x}^{2}S(\frac{n}{x})$
   写到这就完了.

方法二.$\varphi$卷积

根据公式:

$gcd(i,j)={\sum }_{d|gcd(i,j)}\varphi (d)={\sum }_{d|i,d|j}\varphi (d)$

因此

${\sum }_{i=1}^n{\sum }_{j=1}^{n}ijgcd(i,j)={\sum }_{i=1}^n{\sum }_{j=1}^{n}ij{\sum }_{d|i,d|j}\varphi (d)={\sum }_{d=1}^n\varphi (d)d^2{\sum }_{i=1}^{n/d}{\sum }_{j=1}^{n/d}ij={\sum }_{d=1}^n\varphi (d)d^2(\frac{\lfloor \frac{n}{d}\rfloor (\lfloor \frac{n}{d}\rfloor +1)}{2}{)}^2$

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$\varphi$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;
}