[BZOJ 4916]神犇和蒟蒻

传送门

Description

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

求(sum_G)和(sum_F)

Solution 

• For all cases,(G(n)=n)

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

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

• [egin{equation} egin{split} &sum_{i=1}^nsum_{d|i}d cdot phi(d)cdotfrac{i}{d}\=&sum_{i=1}^nicdot sum_{d|i} 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} ]

• [egin{equation} egin{split} Sum_F(n)=&sum_{i=1}^n(Id*F)(i)-sum_{i=2}^{n}icdot F(lfloorfrac{n}{i} floor)\ =&frac{n(n+1)(2n+1)}{6}-sum_{i=2}^{n}icdot F(lfloorfrac{n}{i} floor) 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
",calc(N));
	return 0;
}

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