[数论] HDU 6833 A very easy math problem

题目大意

给定正整数 (n,x,k)，求

[sum_{a_1=1}^{n}sum_{a_2=1}^ndotsbsum_{a_x=1}^nleft(prod_{j=1}^xa_j^k ight)fleft(gcdleft(a_1,a_2,dots,a_x ight) ight)cdot gcdleft(a_1,a_2,dots,a_x ight) ]

如果存在正整数 (k(k>1))，并且 (k^2|x)，那么 (f(x)=0)，否则 (f(x)=1)。

(1leq kleq 10^9,1leq xleq 10^9,1leq nleq 2 imes10^5)

题解

注意到 (f(x)) 的定义，显然 (f(x)=mu^2(x))。
那么就是要求

[sum_{a_1=1}^{n}sum_{a_2=1}^ndotsbsum_{a_x=1}^nleft(prod_{j=1}^xa_j^k ight)mu^2left(gcdleft(a_1,a_2,dots,a_x ight) ight)cdot gcdleft(a_1,a_2,dots,a_x ight) ]

首先枚举gcd

[sum_{d=1}^nsum_{a_1=1}^{n}sum_{a_2=1}^ndotsbsum_{a_x=1}^nleft(prod_{j=1}^xa_j^k ight)mu^2left(d ight)cdot d [d=gcdleft(a_1,a_2,dots,a_x ight)] ]

[sum_{d=1}^nmu^2left(d ight)dsum_{a_1=1}^{n}sum_{a_2=1}^ndotsbsum_{a_x=1}^nleft(prod_{j=1}^xa_j^k ight)[d=gcdleft(a_1,a_2,dots,a_x ight)] ]

[sum_{d=1}^nmu^2left(d ight)dsum_{a_1=1}^{n/d}sum_{a_2=1}^{n/d}dotsbsum_{a_x=1}^{n/d}left(prod_{j=1}^x(a_jd)^k ight)[gcdleft(a_1,a_2,dots,a_x ight)=1] ]

[sum_{d=1}^nmu^2left(d ight)dsum_{a_1=1}^{n/d}sum_{a_2=1}^{n/d}dotsbsum_{a_x=1}^{n/d}left(prod_{j=1}^x(a_jd)^k ight)sum_{p|gcd(a_1,a_2,dots,a_x)}mu(p) ]

[sum_{d=1}^nmu^2left(d ight)d^{kx+1}sum_{a_1=1}^{n/d}sum_{a_2=1}^{n/d}dotsbsum_{a_x=1}^{n/d}left(prod_{j=1}^xa_j^k ight)sum_{p|gcd(a_1,a_2,dots,a_x)}mu(p) ]

枚举 (p) 得

[sum_{d=1}^nmu^2left(d ight)d^{kx+1}sum_{p=1}^{n/d}mu(p)sum_{a_1=1}^{n/dp}sum_{a_2=1}^{n/dp}dotsbsum_{a_x=1}^{n/dp}left(prod_{j=1}^x(a_jp)^k ight) ]

[sum_{d=1}^nmu^2left(d ight)d^{kx+1}sum_{p=1}^{n/d}mu(p)p^{kx}sum_{a_1=1}^{n/dp}sum_{a_2=1}^{n/dp}dotsbsum_{a_x=1}^{n/dp}left(prod_{j=1}^xa_j^k ight) ]

由于

[sum_{a_1=1}^{n}sum_{a_2=1}^ndotsbsum_{a_x=1}^nleft(prod_{j=1}^xa_j^k ight)=left(sum_{i=1}^ni^k ight)^x ]

有

[sum_{d=1}^nmu^2left(d ight)d^{kx+1}sum_{p=1}^{n/d}mu(p)p^{kx}left(sum_{i=1}^{n/dp}i^k ight)^x ]

令 (T=dp)，得

[sum_{T=1}^{n}T^{kx}left(sum_{i=1}^{n/T}i^k ight)^xsum_{d|T}mu^2(d)mu(frac{T}{d})d ]

令

[g(T)=sum_{d|T}mu^2(d)mu(frac{T}{d})d ]

则原式变为

[sum_{T=1}^{n}T^{kx}left(sum_{i=1}^{n/T}i^k ight)^xg(T) ]

对于 (Tin [1,n]) 的所有 (g(T))，我们可以在 (O(nlog n)) 的时间预处理出。
最后对于每一组询问，可以在 (O(sqrt n)) 的时间内通过数论分块求出。
设 (t) 为数据组数，则时间复杂度为 (O(nlog n+tsqrt n))。

Code

#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#include <vector>
using namespace std;

#define RG register int
#define LL long long

template<typename elemType>
inline void Read(elemType &T){
    elemType X=0,w=0; char ch=0;
    while(!isdigit(ch)) {w|=ch=='-';ch=getchar();}
    while(isdigit(ch)) X=(X<<3)+(X<<1)+(ch^48),ch=getchar();
    T=(w?-X:X);
}

const LL MOD=1000000007LL;
const int maxn=200000;
LL g[maxn+5],h[maxn+5],Mu[maxn+5];
bool not_Prime[maxn+5];
vector<int> Prime;
LL K,X;
int T,n;

LL ExPow(LL b,LL n){
    LL x=1,Power=b%MOD;
    while(n){
        if(n&1) x=x*Power%MOD;
        Power=Power*Power%MOD;
        n>>=1;
    }
    return x;
}

void Get_Mu(int Len){
    Mu[1]=1;
    not_Prime[1]=1;
    int Size=0;
    for(RG i=2;i<=Len;i++){
        if(!not_Prime[i]){
            Prime.push_back(i);
            Mu[i]=-1;++Size;
        }
        for(RG j=0;j<Size;j++){
            int mid=Prime[j]*i;
            if(mid>Len) break;
            not_Prime[mid]=1;
            if(i%Prime[j]==0){
                Mu[i*Prime[j]]=0;
                break;
            }
            Mu[mid]=-Mu[i];
        }
    }
    return;
}

void Init(){
    Get_Mu(maxn);
    for(RG i=1;i<=maxn;++i)
        for(RG j=i;j<=maxn;j+=i)
            g[j]=((g[j]+Mu[i]*Mu[i]*Mu[j/i]*i%MOD)%MOD+MOD)%MOD;
    for(RG i=1;i<=maxn;++i){
        LL temp=ExPow(ExPow(i,K),X);
        g[i]=(g[i-1]+g[i]*temp%MOD)%MOD;
        h[i]=(h[i-1]+ExPow(i,K))%MOD;
    }
    return;
}

LL Solve(){
    LL Res=0;
    for(RG L=1,R;L<=n;L=R+1){
        R=n/(n/L);
        LL temp=ExPow(h[n/L],X);
        Res=((Res+temp*(g[R]-g[L-1])%MOD)%MOD+MOD)%MOD;
    }
    return Res;
}

int main(){
    Read(T);Read(K);Read(X);
    Init();
    while(T--){
        Read(n);
        printf("%lld
",Solve());
    }
    return 0;
}