A Very Easy Math Problem（莫比乌斯反演）

A Very Easy Math Problem

题意：

给定$n,x,k$，求解：
$$\sum _{a_1=1}^n\sum _{a_2=1}^n...\sum _{a_x=1}^n(\prod _{j=1}^x{a}_j^k)f(gcd(a_1,a_2,...,a_x))\ast gcd(a_1,a_2,...,a_x)$$
其中
$$f(x)=\{\begin{array}{l}0,(k^2|x)and(k!=1)\\ 1,else\end{array}$$

Solution:

$$\begin{gathered} \sum _{a_1=1}^n\sum _{a_2=1}^n...\sum _{a_x=1}^n(\prod _{j=1}^x{a}_j^k)f(gcd(a_1,a_2,...,a_x))\ast gcd(a_1,a_2,...,a_x) \\ =\sum _{g=1}^n\sum _{a_1=1}^n\sum _{a_2=1}^n...\sum _{a_x=1}^n(\prod _{j=1}^x{a}_j^k)f(g)\ast g\ast [gcd(a_1,a_2,...,a_x)=g] \\ =\sum _{g=1}^{n}f(g)\ast g\sum _{a_1=1}^n\sum _{a_2=1}^n...\sum _{a_x=1}^n(\prod _{j=1}^x{a}_j^k)[gcd(\frac{a_1}{g},\frac{a_2}{g},...,\frac{a_x}{g})=1] \\ =\sum _{g=1}^{n}f(g)\ast g\sum _{a_1=1}^{\lfloor \frac{n}{g}\rfloor }\sum _{a_2=1}^{\lfloor \frac{n}{g}\rfloor }...\sum _{a_x=1}^{\lfloor \frac{n}{g}\rfloor }(\prod _{j=1}^x{a}_j^k{g}^k)\sum _{d|gcd(a_1,a_2,...a_x)}\mu (d) \\ =\sum _{g=1}^{n}f(g)\ast g\ast g^{xk}\sum _{a_1=1}^{\lfloor \frac{n}{g}\rfloor }{a}_1^k\sum _{a_2=1}^{\lfloor \frac{n}{g}\rfloor }{a}_2^k...\sum _{a_x=1}^{\lfloor \frac{n}{g}\rfloor }{a}_x^k\sum _{d|gcd(a_1,a_2,...a_x)}\mu (d) \\ =\sum _{g=1}^{n}f(g)\ast g\ast g^{xk}\sum _{d=1}^{\lfloor \frac{n}{d}\rfloor }\mu (d)(\sum _{a=1}^{\lfloor \frac{n}{gd}\rfloor }{a}^k{d}^k{)}^x \end{gathered}$$
假设T=gd,那么
$$\begin{gathered} \sum _{g=1}^{n}f(g)\ast g\ast g^{xk}\sum _{d=1}^{\lfloor \frac{n}{d}\rfloor }\mu (d)\ast d^{xk}(\sum _{a=1}^{\lfloor \frac{n}{gd}\rfloor }{a}^k{)}^x \\ =\sum _{g=1}^{n}f(g)\ast g\ast g^{xk}\sum _{g|T}\mu (\frac{T}{g})\ast (\frac{T}{g}{)}^{xk}(\sum _{a=1}^{\lfloor \frac{n}{T}\rfloor }{a}^k{)}^x \\ =\sum _{T=1}^n{T}^{xk}(\sum _{g|T}f(g)\ast g\ast \mu (\frac{T}{g}))(\sum _{a=1}^{\lfloor \frac{n}{T}\rfloor }{a}^k{)}^x \end{gathered}$$
之后前缀和+整除分块就可以求解了。

代码

#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long ll;
const int mod=1e9+7;
const int N=2e5;
int mu[N+5],prime[N+5],notPrime[N+5],cnt=0,is[N+5];
ll sum[N+5];
ll fpow(ll x,ll power)
{
   
    ll ans=1;
    while(power)
    {
   
        if(power&1)ans=ans*x%mod;
        power>>=1;
        x=x*x%mod;
    }
    return ans;
}
void init()
{
   
    mu[1]=is[1]=1;
    for(int i=2;i<=N;i++)
    {
   
        if(!notPrime[i])
        {
   
            prime[cnt++]=i;
            mu[i]=-1;
            is[i]=1;
        }
        for(int j=0;j<cnt&&1ll*i*prime[j]<=N;j++)
        {
   
            notPrime[i*prime[j]]=1;
            if(i%prime[j]==0)break;
            mu[i*prime[j]]=-mu[i];
            is[i*prime[j]]=is[i];
        }
    }
}
int t,k,x;
ll a[200005];

int main()
{
   
    init();
    scanf("%d%d%d",&t,&k,&x);
    for(int i=1;i<=N;i++)a[i]=(a[i-1]+fpow(i,k))%mod;
    for(int i=1;i<=N;i++)a[i]=fpow(a[i],x);
    for(int i=1;i<=N;i++)
    {
   
        for(int j=1;j<=N/i;j++)
            sum[i*j]=(sum[i*j]+1ll*j*is[j]*mu[i]%mod)%mod;
    }
    for(int i=1;i<=N;i++)sum[i]=(sum[i]*fpow(i,1ll*k*x)%mod+sum[i-1])%mod;
    while(t--)
    {
   
        int n;scanf("%d",&n);
        ll res=0;
        for(int l=1,r;l<=n;l=r+1)
        {
   
            r=n/(n/l);
            res=(res+1ll*(sum[r]-sum[l-1])*a[n/l]%mod)%mod;
        }
        res=(res+mod)%mod;
        printf("%lld\n",res);
    }
    return 0;
}