P3768 簡單的數學題(杜教篩+莫比烏斯反演)

題意

給定n和p(p爲質數)求$\sum_{i=1}^n\sum_{j=1}^nijgcd(i,j)\% p$
$n\le 10^{10},5\times 10^8\le p\le1.1 \times 10^9$

思路

經典的先提出gcd
$=\sum_{d=1}^nd\sum_{i=1}^n\sum_{j=1}^nij[gcd(i,j)==d]$
後面的部分除以d，有
$=\sum_{d=1}^nd\sum_{i=1}^{\frac{n}{d}}\sum_{j=1}^{\frac{n}{d}}di\ dj[gcd(i,j)==1]$
提出後面的d
$=\sum_{d=1}^nd^3\sum_{i=1}^{\frac{n}{d}}\sum_{j=1}^{\frac{n}{d}}ij[gcd(i,j)==1]$
然後進行莫比烏斯反演，將$\sum_{d|n}\mu(d)=[n==1]$帶入有
$=\sum_{d=1}^nd^3\sum_{i=1}^{\frac{n}{d}}\sum_{j=1}^{\frac{n}{d}}ij\sum_{k|gcd(i,j)}\mu(k)$
將k提出有,及枚舉1到$\frac{n}{d}$內k的倍數
$=\sum_{d=1}^nd^3\sum_{k=1}^{\frac{n}{d}}\mu(k)\sum_{i=1}^{\frac{n}{dk}}\sum_{j=1}^{\frac{n}{dk}}ki\ kj$
$=\sum_{d=1}^nd^3\sum_{k=1}^{\frac{n}{d}}\mu(k)k^2\sum_{i=1}^{\frac{n}{dk}}\sum_{j=1}^{\frac{n}{dk}}ij$
上式等價於
$=\sum_{d=1}^nd^3\sum_{dk=1}^{n}\mu(k)k^2\sum_{i=1}^{\frac{n}{dk}}\sum_{j=1}^{\frac{n}{dk}}ij$
令$T=dk$有
$$
$=\sum_{d=1}^nd^3\sum_{T=1}^{n}[T\% d==0]\mu(\frac{T}{d})(\frac{T}{d})^2\sum_{i=1}^{\frac{n}{T}}\sum_{j=1}^{\frac{n}{T}}ij$
我們將$d^3$乘入
$=\sum_{d=1}^n\sum_{T=1}^{n}[T\% d==0]\mu(\frac{T}{d})dT^2\sum_{i=1}^{\frac{n}{T}}\sum_{j=1}^{\frac{n}{T}}ij$
這裏可以看作枚舉的是1到n中每一個d的倍數，那麼我們也等價於枚舉1到n中每個數T的因子
$$
$=\sum_{T=1}^nT^2\sum_{d|T}d\mu(\frac{T}{d})\sum_{i=1}^{\frac{n}{T}}\sum_{j=1}^{\frac{n}{T}}ij$
對於式子$\sum_{i=1}^{\frac{n}{T}}\sum_{j=1}^{\frac{n}{T}}ij$
有
$\sum_{i=1}^{\frac{n}{T}}i\sum_{j=1}^{\frac{n}{T}}j=\sum_{i=1}^{\frac{n}{T}}i(\frac{(\left \lfloor \frac{n}{T}\right \rfloor+1)\left \lfloor\frac{n}{T}\right \rfloor}{2})=(\frac{(\left \lfloor \frac{n}{T}\right \rfloor+1)\left \lfloor\frac{n}{T}\right \rfloor}{2})^2$
對於式子$\sum_{d|T}d\mu(\frac{T}{d})$可以發現是$id$於$\mu$的狄利克雷卷積$id*\mu=\varphi$
所以有
$=\sum_{T=1}^nT^2\varphi(T)(\frac{(\left \lfloor \frac{n}{T}\right \rfloor+1)\left \lfloor\frac{n}{T}\right \rfloor}{2})^2$
然後可以對
$\sum_{T=1}^nT^2\varphi(T)$用杜教篩
令$f(x)=x^2\varphi(x)$，$g(x)=x^2$，$S(n)=\sum_{i=1}^nf(i)$
$$
$\sum_{i=1}^ng*f=\sum_{i=1}^n\sum_{d|i}g(d)f(\frac{i}{d})=\sum_{i=1}^n\sum_{d|i}d^2(\frac{i}{d})^2\varphi(\frac{i}{d})$
$=\sum_{i=1}^ni^2\sum_{d|i}\varphi(\frac{i}{d})$
有$\sum_{d|n}\varphi(d)=n$，故
$=\sum_{i=1}^ni^3$
再看$\sum_{i=1}^n\sum_{d|i}g(d)f(\frac{i}{d})$
考慮枚舉每個因子的倍數
$\sum_{i=1}^n\sum_{d|i}g(d)f(\frac{i}{d})=\sum_{d=1}^ng(d)\sum_{i=1}^{\left \lfloor\frac{n}{d}\right \rfloor}f(i)=\sum_{d=1}^ng(d)S(\left \lfloor\frac{n}{d}\right \rfloor)$
$g(1)S(n)+\sum_{d=2}^ng(d)S(\left \lfloor\frac{n}{d}\right \rfloor)=\sum_{i=1}^ni^3$
$g(x)=x^2$帶入移項得
$S(n)=\sum_{i=1}^ni^3-\sum_{d=2}^nd^2S(\left \lfloor\frac{n}{d}\right \rfloor)$
有
$\sum_{i=1}^ni^3=(\frac{n(n+1)}{2})^2,\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}{6}$
用杜教篩就可以做了
注意n大於int，預處理時可能會爆，注意取模

#include <bits/stdc++.h>
using namespace std;
const int N=1e7+5;
int prime[N];
long long phi[N];
int vis[N];
int cnt;
long long mod;
long long inv2;
long long inv4;
long long inv6;
long long quickmod(long long a,long long b)
{
    long long ans=1;
    while(b)
    {
        if(b%2==1)
            ans=ans*a%mod;
        b=b/2;
        a=a*a%mod;
    }
    return ans;
}
void init()
{
    cnt=0;
    phi[1]=1;
    for(int i=2;i<N;i++)
    {
        if(!vis[i])
        {
            prime[cnt++]=i;
            phi[i]=i-1;
        }
        for(int j=0;j<cnt&&i*prime[j]<N;j++)
        {
            vis[i*prime[j]]=1;
            if(i%prime[j]==0)
            {
                phi[i*prime[j]]=phi[i]*prime[j];
                break;
            }
            else
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
        }
    }
    for(int i=1;i<N;i++)
        phi[i]=(phi[i-1]+1ll*i*i%mod*phi[i]%mod)%mod;
}
map<long long,long long>P;
long long Sum2(long long n)
{
    return n%mod*((n+1)%mod)%mod*(2*n%mod+1)%mod*inv6%mod;
}
long long Sum(long long n)
{
    if(n<N) return phi[n];
    if(P[n]) return P[n];
    long long sum=n%mod*((n+1)%mod)%mod*(n%mod)%mod*((n+1)%mod)%mod*inv4%mod;
    for(long long i=2,last;i<=n;i=last+1)
    {
        last=n/(n/i);
        sum=(sum-(Sum2(last)-Sum2(i-1)+mod)%mod*Sum(n/i)%mod+mod)%mod;
    }
    return P[n]=sum;
}

int main()
{
    long long n;
    scanf("%lld%lld",&mod,&n);
    init();
    inv2=quickmod(2,mod-2);
    inv4=quickmod(4,mod-2);
    inv6=quickmod(6,mod-2);
    long long sum=0;
    for(long long i=1,last;i<=n;i=last+1)
    {
        last=n/(n/i);
        sum=(sum+(Sum(last)-Sum(i-1)+mod)%mod*((n/i+1)%mod)%mod*((n/i)%mod)%mod*((n/i+1)%mod)%mod*((n/i)%mod)%mod*inv4%mod)%mod;
    }
    printf("%lld\n",sum);
    return 0;
}