洛谷P3768 简单的数学题

洛谷P3768 简单的数学题

题目大意

给出$n$和质数$p$，求
$$\left(\sum _{i=1}^n\sum _{j=1}^{n}ij\gcd (i,j)\right)\mathrm{mod}p$$

题解

前置知识：杜教筛

原式为

$$\sum _{i=1}^n\sum _{j=1}^{n}ij\gcd (i,j)$$

枚举$\gcd$可得

$$\sum _{d=1}^n\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{d}\rfloor }{d}^{3}ij\times [\gcd (i,j)=1]$$

利用莫比乌斯函数的性质，可得

$$\sum _{d=1}^n\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{d}\rfloor }{d}^{3}ij\times \sum _{k|i,k|j}\mu (k)$$

枚举$k$，得

$$\sum _{d=1}^n\sum _{k=1}^{\lfloor \frac{n}{d}\rfloor }\mu (k)\sum _{i=1}^{\lfloor \frac{n}{kd}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{kd}\rfloor }{d}^3{k}^{2}ij$$

稍作变换，可得

$$\sum _{d=1}^n{d}^3\sum _{k=1}^{\lfloor \frac{n}{d}\rfloor }\mu (k)\times k^2(\sum _{i=1}^{\lfloor \frac{n}{kd}\rfloor }i)(\sum _{j=1}^{\lfloor \frac{n}{kd}\rfloor }j)$$

设$sum(n)=\sum _{i=1}^{n}i=\frac{n(n+1)}{2}$，则原式变为

$$\sum _{d=1}^n{d}^3\sum _{k=1}^{\lfloor \frac{n}{d}\rfloor }\mu (k)\times k^2\times sum(\lfloor \frac{n}{kd}\rfloor {)}^2$$

令$T=kd$，枚举$T$，得

$$\sum _{T=1}^{n}sum(\lfloor \frac{n}{T}\rfloor {)}^2\times T^2\sum _{d|T}d\mu (\frac{T}{d})$$

由狄利克雷卷积可得，$\mu \ast Id=\varphi$，于是原式变为

$$\sum _{T=1}^{n}sum(\lfloor \frac{n}{T}\rfloor {)}^2\times T^2\varphi (T)$$

我们可以用数论分块求解，需要$T^2\varphi (T)$的前缀和，用杜教筛来求，步骤如下

由欧拉函数的性质，$n=\sum _{d|n}\varphi (\frac{n}{d})$

那么

$$n^3=n^2\sum _{d|n}\varphi (d)=\sum _{d|n}(\frac{n}{d}{)}^2\varphi (\frac{n}{d})\times d^2=n^2\varphi (n)+\sum _{d|n,d>1}(\frac{n}{d}{)}^2\varphi (\frac{n}{d})\times d^2$$

由此可得

$$n^2\varphi (n)=n^3-\sum _{d|n,d>1}(\frac{n}{d}{)}^2\varphi (\frac{n}{d})\times d^2$$

设$f(T)=T^2\varphi (T)$，$F$为$f$的前缀和，则

$$f(n)=n^3-\sum _{d|n,d>1}{d}^2\times f(\frac{n}{d})$$

分别求前缀和得

$$F(n)=\sum _{i=1}^n{i}^3-\sum _{i=1}^n\sum _{d|i,d>1}{d}^2\times f(\frac{i}{d})$$

枚举$d$得

$$F(n)=\sum _{i=1}^n{i}^3-\sum _{d=2}^n{d}^2\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }f(i)=\sum _{i=1}^n{i}^3-\sum _{d=2}^n{d}^{2}F(\lfloor \frac{n}{d}\rfloor )$$

注：

• ${\sum }_{i=1}^n{i}^3=(\sum _{i=1}^{n}i{)}^2$
• $\sum _{i=1}^n{i}^2=\frac{n(n+1)(2n+1)}{6}$

code

#include
using namespace std;
const int N=5000000;
int z[N+5],pr[N+5];
long long p,ans=0,ny,ph[N+5],sph[N+5];
map<long long,long long>mph;
long long mi(long long t,long long v){
	if(v==0) return 1;
	long long re=mi(t,v/2);
	re=re*re%p;
	if(v&1) re=re*t%p;
	return re; 
}
void init(){
	ph[1]=1;
	for(int i=2;i<=N;i++){
		if(!z[i]){
			pr[++pr[0]]=i;ph[i]=i-1;
		}
		for(int j=1;j<=pr[0]&&i*pr[j]<=N;j++){
			z[i*pr[j]]=1;
			if(i%pr[j]==0){
				ph[i*pr[j]]=ph[i]*pr[j];
				break;
			}
			ph[i*pr[j]]=ph[i]*(pr[j]-1);
		}
	}
	for(int i=1;i<=N;i++){
		sph[i]=(sph[i-1]+ph[i]*i%p*i%p)%p;
	}
}
long long sum(long long t){
	t=t%p;
	return t*(t+1)/2%p;
}
long long tw(long long t){
	t=t%p;
	return t*(t+1)%p*(2*t+1)%p*ny%p;
}
long long djs(long long n){
	if(n<=N) return sph[n];
	if(mph.count(n)) return mph[n];
	long long t,re=0;
	for(long long l=2,r;l<=n;l=r+1){
		r=n/(n/l);
		t=(tw(r)-tw(l-1)+p)%p;
		re=(re+t*djs(n/l)%p)%p;
	}
	t=sum(n);
	re=(t*t%p-re+p)%p;
	return mph[n]=re;
}
int main()
{
	long long n;
	scanf("%lld%lld",&p,&n);
	init();
	ny=mi(6ll,p-2);
	for(long long l=1,r;l<=n;l=r+1){
		r=n/(n/l);
		long long t=sum(n/l);
		ans=(ans+(djs(r)-djs(l-1)+p)%p*t%p*t%p)%p;
	}
	printf("%lld",ans);
	return 0;
}