洛谷P3768 简单的数学题解题报告

$$\begin{eqnarray}&\sum_{i=1}^{n}\sum_{j=1}^{n}ij\gcd(i,j)\\&\sum_{d=1}^{n}\sum_{i=1}^{n}\sum_{j=1}^{n}ij[\gcd(i,j)=d] \\&\sum_{d=1}^{n}d^3\sum_{i=1}^{\frac{n}{d}}\sum_{j=1}^{\frac{n}{d}}ij[\gcd(i,j)=1] \\&\sum_{d=1}^{n}d^3\sum_{i=1}^{\frac{n}{d}}\sum_{j=1}^{\frac{n}{d}}ij\sum_{x|\gcd(i,j)}\mu(x) \\&\sum_{d=1}^{n}d^3\sum_{x=1}^{\frac{n}{d}}x^2\mu(x)\sum_{i=1}^{\frac{n}{dx}}\sum_{j=1}^{\frac{n}{dx}}ij \\&\sum_{d=1}^{n}d^3\sum_{x=1}^{\frac{n}{d}}x^2\mu(x)(1+2+3+…\lfloor \frac{n}{xd} \rfloor )^2 \\&令s(x)=(1+x)*x/2 \\&\sum_{d=1}^{n}d^3\sum_{x=1}^{\frac{n}{d}}x^2\mu(x)s(\lfloor\frac{n}{xd}\rfloor)^2\\&令T=dx \\&\sum_{T=1}s(\lfloor\frac{n}{T}\rfloor)^2\sum_{d|T}d^3\frac{T}{d}^2\mu(\frac{T}{d})\\&\sum_{T=1}s(\lfloor\frac{n}{T}\rfloor)^2T^2\sum_{d|T}d\mu(\frac{T}{d})\\&\sum_{T=1}s(\lfloor\frac{n}{T}\rfloor)^2T^2\varphi(T)\\&令f(x)=x^2\varphi(x)\\&sum(n)=\sum_{i=1}^{n}(g*f)(i)-\sum_{i=2}^{n}g(i)sum(n/i)(杜教筛式子)\\&(g*f)(i)=i*i\sum_{d|i}\varphi(d)=i^3\\&sum(n)=\sum_{i=1}^{n} i^{3}-\sum_{i=2}^{n} i^{2} sum\left(\frac{n}{i}\right)\\&ans=\sum_{T=1}^{n} \operatorname{sum}\left(\frac{n}{T}\right)^{2} T^{2} \sum_{d|T} d \mu\left(\frac{T}{d}\right)\\&\end{eqnarray}$$

#include<bits/stdc++.h>
#define int long long
using namespace std;
const long long INF=1LL<<31;
int N=8000000;
int cnt,n;
long long p[8001000],inv2,inv6,ans,zhi[8001000],mod;
bool he[8001000];
map<long long,long long>M;
long long S(long long x){x%=mod;return x*(x+1)%mod*inv2%mod;}
long long Sump(long long x){x%=mod;return x*(x+1)%mod*(x+x+1)%mod*inv6%mod;}
void xxs()
{
    he[1]=p[1]=1;
    for(int i=2;i<=N;i++)
    {
        if(he[i]==0)
        {
            p[i]=(i-1)%mod;
            zhi[++cnt]=i;
        }
        for(int j=1;j<=cnt&&i*zhi[j]<=N;j++)
        {
            he[i*zhi[j]]=true;
            if(i%zhi[j]==0)
            {
                p[i*zhi[j]]=1LL*p[i]*zhi[j]%mod;
                break;
            }
            else
            {
                p[i*zhi[j]]=1LL*p[i]*(zhi[j]-1)%mod;
            }
        }
    }
    for(int i=1;i<=N;i++)p[i]=(p[i-1]+1ll*p[i]*i%mod*i%mod)%mod;
}
long long SF(long long x)
{
    if(x<=N)
        return p[x];
    if(M.find(x)!=M.end())
        return M[x];
    long long ret=S(x);
    ret=ret*ret%mod;
    for(long long i=2,r;i<=x;i=r+1)
    {
        r=x/(x/i);
        long long tt=(Sump(r)-Sump(i-1))%mod;
        ret-=SF(x/i)*tt%mod;
        ret%=mod;
    }
    return M[x]=(ret+mod)%mod;
}
long long quick_pow(long long a,long long b)
{
    long long res=1;
    while(b>0)
    {
        if(b&1)
        {
            res*=a;
            res%=mod;
        }
        a*=a;
        a%=mod;
        b>>=1;
    }
    return res;
}
signed main()
{
    scanf("%lld%lld",&mod,&n);
    inv2=quick_pow(2,mod-2);
    inv6=quick_pow(6,mod-2);
    xxs();
    for(long long i=1,r;i<=n;i=r+1)
    {
        r=n/(n/i);
        long long tt=S(n/i);
        tt=tt*tt%mod;
        long long gg=(SF(r)-SF(i-1))%mod;
        ans+=gg*tt%mod;
        ans%=mod;
    }
    printf("%lld\n",(ans+mod)%mod);
    return 0;
}