51nod 1238 最小公倍数之和 V3 【欧拉函数+杜教筛】

首先题目中给出的代码打错了，少了个等于号，应该是

G=0;
for(i=1;i<=N;i++)
for(j=1;j<=N;j++)
{
    G = (G + lcm(i,j)) % 1000000007;
}

然后就是大力推公式：
\[ \sum_{i=1}^{n}\sum_{j=1}^{n}lcm(i,j) \]
\[ =\sum_{i=1}^{n}\sum_{j=1}^{n}\frac{ij}{gcd(i,j)} \]
\[ =\sum_{d=1}^{n}\sum_{i=1}^{n}\sum_{j=1}^{n}[gcd(i,j)==d]\frac{ij}{d} \]
\[ =\sum_{d=1}^{n}\sum_{i=1}^{\left \lfloor \frac{n}{d} \right \rfloor}\sum_{j=1}^{\left \lfloor \frac{n}{d} \right \rfloor}[gcd(i,j)==d]ijd \]
然后需要一个打表找规律，发现\( \sum_{i=1}^{n}\sum_{j=1}^{i}[gcd(i,j)==1]ij=\sum_{i=1}^{n}i\frac{i\phi(i)+[i==1]}{2} \)
\[ =\sum_{d=1}^{n}d((2)\sum_{i=1}^{\left \lfloor \frac{n}{d} \right \rfloor}i\frac{i\phi(i)+[i==1]}{2})-1) \]
\[ =\sum_{d=1}^{n}d((\sum_{i=1}^{\left \lfloor \frac{n}{d} \right \rfloor}i^2\phi(i)+[i==1])-1) \]
\[ =\sum_{d=1}^{n}d\sum_{i=1}^{\left \lfloor \frac{n}{d} \right \rfloor}i^2\phi(i) \]
然后就可以递归使用杜教筛了，关于用杜教筛求\( \sum_{i=1}^{n}i^2\phi(i) \)的前缀和，有如下推导：
设
\[ g(n)=\sum_{i=1}^{n}i^2\sum_{d=1}^{i}[d|i]\phi(d)=\sum_{i=1}^{n}i^3=\frac{n^2(n+1)^2}{4} \]
\[ s(n)=\sum_{i=1}^{n}i^2\phi(i) \]那么把g展开：
\[ g(n)=\sum_{i=2}^{n}i^2\sum_{d=1}^{i-1}[d|i]\phi(d)+s(n) \]
\[ s(n)=g(n)-\sum_{i=2}^{n}i^2\sum_{d=1}^{i-1}[d|i]\phi(d) \]
\[ =g(n)-\sum_{k=2}^{n}k^2\sum_{d=1}^{\left \lfloor \frac{n}{k} \right \rfloor}d\phi(d) \]
\[ =g(n)-\sum_{k=2}^{n}k^2*s(\left \lfloor \frac{n}{k} \right \rfloor) \]
\[ =\frac{n^2(n+1)^2}{4}-\sum_{k=2}^{n}k^2*s(\left \lfloor \frac{n}{k} \right \rfloor) \]
这就是标准的杜教筛递归子问题形式了，直接求解即可。

#include
#include
using namespace std;
const long long N=1000005,m=1000000,inv2=500000004,inv4=250000002,inv6=166666668,mod=1e9+7;
long long n,phi[N],q[N],tot,ans,ha[N];
bool v[N];
long long wk1(long long x)
{
    if(x>=mod)
        x-=mod;
    return x%mod*(x+1)%mod*inv2%mod;
}
long long wk2(long long x)
{
    if(x>=mod)
        x-=mod; 
    return x%mod*(x+1)%mod*(x%mod*2+1)%mod*inv6%mod;
}
long long wk3(long long x)
{
    if(x>=mod)
        x-=mod;
    return x%mod*x%mod*(x+1)%mod*(x+1)%mod*inv4%mod;
}
long long slv(long long x)
{
    if(x<=m)
        return phi[x];
    if(ha[n/x])
        return ha[n/x];
    long long re=wk3(x);
    for(long long i=2,la;i<=x;i=la+1)
    {
        la=x/(x/i);
        re=(re-(wk2(la)-wk2(i-1))%mod*slv(x/i)%mod)%mod;
    }
    return ha[n/x]=re;
}
int main()
{
    phi[1]=1;
    for(int i=2;i<=m;i++)
    {
        if(!v[i])
        {
            q[++tot]=i;
            phi[i]=i-1;
        }
        for(int j=1;j<=tot&&i%mod*q[j]<=m;j++)
        {
            int k=i%mod*q[j];
            v[k]=1;
            if(i%q[j]==0)
            {
                phi[k]=phi[i]%mod*q[j];
                break;
            }
            phi[k]=phi[i]%mod*(q[j]-1);
        }
    }
    for(int i=1;i<=m;i++)
        phi[i]=(phi[i]%mod*i%mod*i%mod+phi[i-1])%mod;
    scanf("%lld",&n);
    for(long long i=1,la;i<=n;i=la+1)
    {
        la=n/(n/i);
        ans=(ans+(wk1(la)-wk1(i-1))%mod*slv(n/i)%mod)%mod;
    }
    printf("%lld\n",(ans%mod+mod)%mod);
    return 0;
}

转载于:https://www.cnblogs.com/lokiii/p/8334512.html