HDU 6833 A Very Easy Math Problem（莫比乌斯反演）

原题题面

Given you n,x,k , find the value of the following formula:
${\sum }_{a_1=1}^n{\sum }_{a_2=1}^n...{\sum }_{a_x=1}^n({\prod }_{j=1}^x{a}_j^k)f(gcd(a_1,a_2,...a_x))\ast gcd(a_1,a_2,...a_x)$
$gcd(a_1,a_2,\dots ,a_n)$ is the greatest common divisor of $a_1,a_2,...,a-n$.
The function $f(x)$ is defined as follows:
If there exists an ingeter $k$ $(k>1)$ , and $k^2$ is a divisor of $x$,
then $f(x)=0$, else $f(x)=1$.

输入格式

The first line contains three integers $t,k,x(1\le t\le 10^4,1\le k\le 10^9,1\le x\le 10^9)$
Then $t$ test cases follow. Each test case contains an integer $n$ $(1\le n\le 2\times 10^5)$

输出格式

For each test case, print one integer — the value of the formula.
Because the answer may be very large, please output the answer modulo $10^9+7$.

输入样例

3 1 3
56
5
20

输出样例

139615686
4017
11554723

题面分析

设$gcd(a_1,a_2,...a_x)=d$，得
${\sum }_{a_1=1}^n{\sum }_{a_2=1}^n...{\sum }_{a_x=1}^n({\prod }_{j=1}^x{a}_j^k)f(d)\ast d[gcd(a_1,a_2,...a_x)=d]$
枚举$d$，得到
${\sum }_{d=1}^{n}d\ast f(d)\ast {\sum }_{a_1=1}^n{\sum }_{a_2=1}^n...{\sum }_{a_x=1}^n({\prod }_{j=1}^x{a}_j^k)[gcd(a_1,a_2,...a_x)=d]$
${\sum }_{d=1}^n{d}^{kx+1}\ast f(d)\ast {\sum }_{a_1=1}^{\lfloor \frac{n}{d}\rfloor }{\sum }_{a_2=1}^{\lfloor \frac{n}{d}\rfloor }...{\sum }_{a_x=1}^{\lfloor \frac{n}{d}\rfloor }({\prod }_{j=1}^x{a}_j^k)[gcd(a_1,a_2,...a_x)=1]$
将$[gcd(a_1,a_2,...a_x)=1]$化作$\mu$，得到
${\sum }_{d=1}^n{d}^{kx+1}\ast f(d)\ast {\sum }_{a_1=1}^{\lfloor \frac{n}{d}\rfloor }{\sum }_{a_2=1}^{\lfloor \frac{n}{d}\rfloor }...{\sum }_{a_x=1}^{\lfloor \frac{n}{d}\rfloor }({\prod }_{j=1}^x{a}_j^k){\sum }_{t|a_1,t|a_2,...t|a_x}\mu (t)$
${\sum }_{d=1}^n{d}^{kx+1}\ast f(d)\ast ({\sum }_{i=1}^{\lfloor \frac{n}{d}\rfloor }{i}^k{)}^x\ast {\sum }_{t|a_1,t|a_2,...t|a_x}\mu (t)$
枚举$t$得
${\sum }_{d=1}^n{d}^{kx+1}\ast t^{kx}\ast f(d)\ast {\sum }_{t=1}^{\lfloor \frac{n}{d}\rfloor }\mu (t)({\sum }_{i=1}^{\lfloor \frac{n}{dt}\rfloor }{i}^k{)}^x$
设$T=td$，得到
${\sum }_{T=1}^n{T}^{kx}df(d)\ast ({\sum }_{i=1}^{\lfloor \frac{n}{T}\rfloor }{i}^k{)}^x\ast {\sum }_{t|T}\mu (t)$
为了方便计算，我们把右边的$t|T$改成$d|T$，得到
${\sum }_{T=1}^n{T}^{kx}({\sum }_{i=1}^{\lfloor \frac{n}{T}\rfloor }{i}^k{)}^x\ast {\sum }_{d|T}\mu (d)df(\frac{T}{d})$
设$G(x)={\sum }_{d|T}\mu (d)df(\frac{T}{d})$
首先$\mu$可以$O(n)$预处理，$f$也可以$O(nlogn)$预处理
那么G可以在$O(n)$下预处理
对于$({\sum }_{i=1}^{\lfloor \frac{n}{T}\rfloor }{i}^k{)}^x$，也可以在$O(nlogn)$下处理完
对于每一次查询，利用分块和预处理前缀和，只需要$O(\sqrt{n})$。
故总复杂度为$O(nlogn+T\sqrt{n})$

AC代码(2199ms)

#include <bits/stdc++.h>
using namespace std;
const long long maxn=2e5;
const long long mod=1e9+7;
bool check[maxn+10];//访问标记
int prime[maxn+10];//质数
int mu[maxn+10];//mu函数
int f[maxn+10];//f函数
long long powk[maxn+10];
long long G[maxn+10];
long long sumG[maxn+10];
long long n, k, x;
inline long long quick_pow(long long a, long long b)//快速幂
{
    long long ans=1, base=a;
    while(b!=0)
    {
        if (b&1)
            ans=(long long) ans*base%mod;
        base=(long long) base*base%mod;
        b>>=1;
    }
    return ans;
}
void init()
{
    f[1]=mu[1]=1;
    int tot=0;
    for(int i=2; i<=maxn; i++)
    {
        f[i]=1;
        if (!check[i])
        {
            prime[tot++]=i;
            mu[i]=-1;
        }
        for(int j=0; j<tot; j++)
        {
            if (i*prime[j]>maxn) break;
            check[i*prime[j]]=true;
            if (i%prime[j]==0)
            {
                mu[i*prime[j]]=0;
                break;
            }
            else
            {
                mu[i*prime[j]]=-mu[i];
            }
        }
    }
    for(long long d=2; d<=maxn; d++)//处理f
    {
        for(long long j=d*d; j<=maxn; j+=d*d)
        {
            f[j]=0;
        }
    }
    G[0]=0;
    for(long long d=1; d<=maxn; d++)//处理G
    {
        for(long long i=d; i<=maxn; i+=d)
        {
            G[i]=(G[i]+d*f[d]%mod*mu[i/d]%mod+mod)%mod;
        }
    }
    powk[0]=0;
    sumG[0]=0;
    for(long long i=1; i<=maxn; i++)//处理累加
    {
        long long z=quick_pow(i, k);
        powk[i]=(powk[i-1]+z)%mod;
        sumG[i]=(sumG[i-1]+quick_pow(z, x)*G[i]%mod)%mod;
    }
}
long long answer()
{
    long long ans=0;
    for(long long l=1, r; l<=n; l=r+1)
    {
        r=n/(n/l);
        ans=(ans+(sumG[r]-sumG[l-1]+mod)%mod*quick_pow(powk[n/l], x)%mod)%mod;
    }
    return ans;
}
void solve()
{
    int t;
    scanf("%d%lld%lld", &t, &k, &x);
    init();
    while(t--)
    {
        scanf("%lld", &n);
        printf("%lld\n", answer());
    }
}
int main()
{
//    ios_base::sync_with_stdio(false);
//    cin.tie(0);
//    cout.tie(0);
#ifdef ACM_LOCAL
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    long long test_index_for_debug=1;
    char acm_local_for_debug;
    while(cin>>acm_local_for_debug)
    {
        cin.putback(acm_local_for_debug);
        if (test_index_for_debug>100)
        {
            throw runtime_error("Check the stdin!!!");
        }
        auto start_clock_for_debug=clock();
        solve();
        auto end_clock_for_debug=clock();
        cout<<"\nTest "<<test_index_for_debug<<" successful"<<endl;
        cerr<<"Test "<<test_index_for_debug++<<" Run Time: "
            <<double(end_clock_for_debug-start_clock_for_debug)/CLOCKS_PER_SEC<<"s"<<endl;
        cout<<"--------------------------------------------------"<<endl;
    }
#else
    solve();
#endif
    return 0;
}

DrGilbert 2020.8.7