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Description

共\(T(T\leq3)\)組測試數據。給出\(n,m(n\leq5\times10^5)\)，求
\[ \sum_{i=1}^n \sum_{j=1}^m lcm(i,j)^{gcd(i,j)}\]

Solution

喜聞樂見推推推。
\[\begin{align*} ans &= \sum_{i=1}^n \sum_{j=1}^m lcm(i,j)^{gcd(i,j)} \&= \sum_{d=1}^{+∞} \sum_{i=1}^n \sum_{j=1}^m [gcd(i,j)=d](\frac{ij}{d})^d \&= \sum_{d=1}^{+∞} \sum_{i=1}^{?\frac{n}{d}?} \sum_{j=1}^{?\frac{m}{d}?} [gcd(i,j)=1](ijd)^d \&= \sum_{d=1}^{+∞} d^d \sum_{d_1=1}^{+∞}\mu(d_1) \sum_{d_1|i}^{?\frac{n}{d}?} \sum_{d_1|j}^{?\frac{m}{d}?} (ij)^d \&= \sum_{d=1}^{+∞} d^d \sum_{d_1=1}^{+∞}\mu(d_1) \sum_{i=1}^{?\frac{n}{dd_1}?} \sum_{j=1}^{?\frac{m}{dd_1}?} (id_1\cdot jd_1)^d \&= \sum_{d=1}^{+∞} d^d \sum_{d_1=1}^{+∞}\mu(d_1)d_1^{2d} \sum_{i=1}^{?\frac{n}{dd_1}?} i^d\sum_{j=1}^{?\frac{m}{dd_1}?} j^d \\end{align*}\]

根據調和級數總復雜度為\(O(nlogn)\)。

Code

//DZY Loves Math VI
#include <algorithm>
#include <cstdio>
using std::swap;
typedef long long lint;
const int N=5e5+10;
const int P=1e9+7;
int prCnt,pr[N]; bool prNot[N];
int mu[N];
void init(int n)
{
    mu[1
 
]=1;
    for(int i=2;i<=n;i++)
    {
        if(!prNot[i]) pr[++prCnt]=i,mu[i]=-1;
        for(int j=1;j<=prCnt;j++)
        {
            int x=i*pr[j]; if(x>n) break;
            prNot[x]=true;
            if(i%pr[j]) mu[x]=-mu[i]; else break;
        }
    }
}
int powD[N],S[N];
int pow(int
 
 x,int y)
{
    lint r=1,t=x;
    for(int i=y;i;i>>=1,t=(t*t)%P) if(i&1) r=(r*t)%P;
    return r;
}
int main()
{
    int task=1; init(5e5);
    while(task--)
    {
        int n,m; scanf("%d%d",&n,&m);
        if(n>m) swap(n,m);
        int ans=0;
        for(int i=1;i<=m;i++) powD[i]=1;
        for(int d=1;d<=n;d++)
        {
            int k1=n/d,k2=m/d,ansD=0;
            for(int i=1;i<=k2;i++) powD[i]=(1LL*powD[i]*i)%P;
            for(int i=1;i<=k2;i++) S[i]=(S[i-1]+powD[i])%P;
            for(int d1=1;d1<=k1;d1++) 
            {
                lint t=1LL*(mu[d1]+P)%P*powD[d1]%P*powD[d1]%P;
                t=t*S[k1/d1]%P*S[k2/d1]%P;
                ansD=(ansD+t)%P;
            }
            ans=(ans+1LL*ansD*pow(d,d))%P;
        }
        printf("%d\n",ans);
    }
    return 0;
}

BZOJ3561 - DZY Loves Math VI