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Description

\[ sum_G(n)=\sum_{i=1}^n \mu(i^2)\\sum_F(n)=\sum_{i=1}^n \phi(i^2) \]

Solution?

• For all cases,\(G(n)=n\)

  Because \(G(i^2)=[i==1]\)

• \[ \begin{equation} \begin{split} F(n)&=\sum_{i=1}^n \phi(i^2)\\ &=\sum_{i=1}^{n} i\cdot \phi(i)\\ \end{split} \end{equation}\\ \]

• \[ \begin{equation} \begin{split} &\sum_{i=1}^n\sum_{d|i}d \cdot \phi(d)\cdot\frac{i}ze8trgl8bvbq\\=&\sum_{i=1}^ni\cdot \sum_{d|i}d\cdot\phi(d)\\ =&\sum_{i=1}^ni^2 \\=&\frac{n(n+1)(2n+1)}{6} \end{split} \end{equation} \]

• \[ \sum_{i=1}^n(Id*F)(i)=\frac{n(n+1)(2n+1)}{6} \]

• \[ \begin{equation} \begin{split} Sum_F(n)=&\sum_{i=1}^n(Id*F)(i)-\sum_{i=2}^{n}i\cdot F(\lfloor\frac{n}{i}\rfloor)\\ =&\frac{n(n+1)(2n+1)}{6}-\sum_{i=2}^{n}i\cdot F(\lfloor\frac{n}{i}\rfloor) \end{split} \end{equation} \]

Code?

#include<bits/stdc++.h> #define reg register using namespace std; inline int read() {int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}return x*f; } const int MN=1e9,MX=1e6+5,M=1e6,mod=1e9+7,inv6=166666668,inv2=500000004; int N,F[MX],phi[MX]; int Mul(int x,int y){return 1ll*x*y%mod;} int Add(int x,int y){return (x+y)%mod;} void init() {static int prime[MX],tot;static bool mark[MX];reg int i,j;phi[1]=1;for(i=2;i<=M;++i){if(!mark[i]){prime[++tot]=i;phi[i]=i-1;}for(j=1;j<=tot&&i*prime[j]<=M;++j){mark[i*prime[j]]=true;if(i%prime[j]==0){phi[i*prime[j]]=Mul(phi[i],prime[j]);break;}else phi[i*prime[j]]=Mul(phi[i],phi[prime[j]]);}}for(i=1;i<=M;++i) phi[i]=Add(Mul(phi[i],i),phi[i-1]); } int S(int x){return Mul(x,Mul(x+1,inv2));} int calc(int n) {if(n<=M) return phi[n];if(F[N/n]) return F[N/n];int res=Mul(Mul((n+1),Mul((2*n+1),n)),inv6);for(int l=2,r;l<=n;l=r+1){r=n/(n/l);res=Add(res,mod-Mul(Add(S(r),mod-S(l-1)),calc(n/l)));}return F[N/n]=res; }int main() {N=read();puts("1");init();printf("%lld\n",calc(N));return 0; }

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轉載于:https://www.cnblogs.com/PaperCloud/p/bzoj_4916.html

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