【简单的数学题】

先开始化柿子

求的是

[Ans=sum_{i=1}^Nsum_{j=1}^N i imes j imes gcd(i.j) ]

还是先上套路

[F(n)=sum_{i=1}^Nsum_{j=1}^N[n|(i,j)]i imes j=(frac{(left lfloor frac{N}{n} ight floor+1)left lfloor frac{N}{n} ight floor}{2})^2n^2 ]

[f(n)=sum_{i=1}^Nsum_{j=1}^N[n=(i,j)]i imes j ]

上套路我们要求的是

[Ans=sum_{i=1}^Ni imes f(i)=sum_{i=1}^Nisum_{i|d}mu(frac{d}{i})F(d) ]

[=sum_{d=1}^N(frac{(left lfloor frac{N}{d} ight floor+1)left lfloor frac{N}{d} ight floor}{2})^2sum_{i|d}mu(frac{d}{i}) imes d^2 imes i ]

发现(sum_{i|d}mu(frac{d}{i}) imes i)就是(mu imes id=varphi)

于是就有了

[Ans=sum_{d=1}^N(frac{(left lfloor frac{N}{d} ight floor+1)left lfloor frac{N}{d} ight floor}{2})^2varphi(d)d^2 ]

化到这里就可以直接线筛了，但是这个题的数据范围有点大，需要杜教筛

先把杜教筛的套路拿出来

[S(n)=sum_{i=1}^nh(i)-sum_{i=2}^ng(i)S(left lfloor frac{n}{i} ight floor) ]

这里我们要合理的选取(g,h)

由于

[sum_{i|d}varphi(i)=d ]

这里有了一个(d^2)，尝试给他陪上一个函数使得(d^2)变成一个可以提出来的东西就好了

显然(d^2 imes (frac{n}{d})^2=n^2)

于是设(h(i)=i^2,f(i)=varphi(i)i^2)

我们发现这两个函数卷起来就是

[h(n)=sum_{i|n}varphi(i) imes i^2 imes(frac{n}{i})^2=n^2sum_{i|n}varphi(i)=n^3 ]

有一个好像并不是那么常用的公式

[sum_{i=1}^ni^3=frac{n^2(n+1)^2}{4}=(frac{n(n+1)}{2})^2 ]

[sum_{i=1}^ni^2=frac{n(n+1)(2n+1)}{6} ]

现在的(g,h)都可以快速求了

于是这道题就解决了

#include<iostream>
#include<cstring>
#include<cstdio>
#include<tr1/unordered_map>
#define re register
#define maxn 5000005
#define LL long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
using namespace std::tr1;
unordered_map<int,LL> ma;
LL n,M;
int f[maxn],p[maxn>>1];
LL phi[maxn];
LL P,inv_6,inv_2;
LL exgcd(LL a,LL b,LL &x,LL &y)
{
    if(!b) return x=1,y=0,a;
    LL r=exgcd(b,a%b,y,x);
    y-=a/b*x;
    return r;
}
LL quick(LL a,LL b)
{
    LL S=1;
    while(b){if(b&1ll) S=S*a%P;b>>=1ll;a=a*a%P;}
    return S; 
}
inline LL S(LL x) {x%=P;return x*(x+1)%P*(x+x+1)%P*inv_6%P;}
inline LL Sum(LL x){x%=P;return x*(x+1)%P*inv_2%P;}
LL solve(LL x)
{
    if(x<=M) return phi[x];
    if(ma.find(x)!=ma.end()) return ma[x];
    LL ans=Sum(x)%P;
    ans=ans*ans%P;
    for(re LL l=2,r;l<=x;l=r+1)
    {
        r=x/(x/l);
        ans=(ans-(S(r)-S(l-1)+P)%P*solve(x/l)%P)%P;
    }
    return ma[x]=(ans%P+P)%P;
}
signed main()
{
    scanf("%lld%lld",&P,&n);
    LL x,y;
    inv_2=quick(2ll,P-2);
    inv_6=quick(6ll,P-2);
    M=min(5000000,n);
    f[1]=1,phi[1]=1;
    for(re LL i=2;i<=M;i++)
    {
        if(!f[i]) p[++p[0]]=i,phi[i]=i-1;
        for(re int j=1;j<=p[0]&&p[j]*i<=M;j++)
        {
            f[p[j]*i]=1;
            if(i%p[j]==0)
            {
                phi[p[j]*i]=p[j]*phi[i];
                break;
            }
            phi[p[j]*i]=phi[p[j]]*phi[i];
        }
    }
    for(re LL i=1;i<=M;i++) phi[i]=(phi[i-1]+phi[i]*(LL)i%P*(LL)i)%P;
    LL ans=0;
    for(re LL l=1,r;l<=n;l=r+1)
    {
        r=n/(n/l);
        LL s=Sum(n/l)%P;
        s=s*s%P;
        ans=(ans+s*(((solve(r)-solve(l-1)%P)+P)%P)%P)%P;
    }
    printf("%lld
",ans);
    return 0;
}