洛谷3768 简单的数学题

Problem

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

Solution

为了方便表达，下文中约定$S_1(n)=1+2+\cdots +n$

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$$\sum _{d=1}^{n}d\sum _{i=1}^n\sum _{j=1}^{n}ij[\gcd (i,j)=d]\text{\hspace{1em}(1)}$$

$$\sum _{d=1}^n{d}^3\sum _{i=1}^{n/d}\sum _{j=1}^{n/d}ij[\gcd (i,j)=1]\text{\hspace{1em}(2)}$$

考虑把后面的式子莫反。设$F(n)={\sum }_{i=1}^x{\sum }_{j=1}^{x}ij[\gcd (i,j)=n]$，构造
$$G(n)=\sum _{n|d}F(d)=(\sum _{n|d}d{)}^2=n^2{S}_1(\frac{x}{d}{)}^2\text{\hspace{1em}(3)}$$

反演得
$$F(1)=\sum _{i=1}^x\mu (i)G(i)\text{\hspace{1em}(4)}$$
把$(4)$代入$(2)$中，式子变为
$$\sum _{d=1}^n{d}^3\sum _{i=1}^{n/d}\mu (i)i^2{S}_1(\frac{n}{id}{)}^2\text{\hspace{1em}(5)}$$

$$\sum _{id=1}^n{S}_1(\frac{n}{id}{)}^{2}i{d}^2\phi (id)\text{\hspace{1em}(6)}$$

前面数论分块，后面是一个积性函数，注意到$n\le 10^{10}$，考虑杜教筛

不妨令后面的函数为$f(n)=n^2\phi (n)$，那么我们构造一个$g$。考虑它们的狄利克雷卷积$(f\ast g)(n)={\sum }_{d|n}{d}^2\phi (d)g(\frac{n}{d})$

不妨令$g(n)=n^2$，就可以消掉$d$了，即$(f\ast g)(n)=n^3$。

套用式子即可。$S(n)=\frac{(n(n+1){)}^2}{4}-{\sum }_{i=2}^{n}g(i)S(n/i)$

Code

#include 
#include 
using namespace std;
typedef long long ll;
const int maxn=10000010,N=10000000;
template <typename Tp> inline int getmin(Tp &x,Tp y){return y<x?x=y,1:0;}
template <typename Tp> inline int getmax(Tp &x,Tp y){return y>x?x=y,1:0;}
template <typename Tp> inline void read(Tp &x)
{
    x=0;int f=0;char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
    if(ch=='-') f=1,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    if(f) x=-x;
}
int mod,inv6,ans,tot,pri[1000010],phi[maxn];
ll n;
map<ll,int> ma;
map<ll,int>::iterator itr;
int pls(int x,int y){return x+y>=mod?x+y-mod:x+y;}
int dec(int x,int y){return x-y<0?x-y+mod:x-y;}
int sum(ll x){if(x>=mod) x%=mod;return (ll)x*(x+1)/2%mod;}
int sum2(ll x){if(x>=mod) x%=mod;return (ll)x*(x+1)%mod*(x+x+1)%mod*inv6%mod;}
int power(int x,int y)
{
	int res=1;
	for(;y;y>>=1,x=(ll)x*x%mod)
	  if(y&1)
	    res=(ll)res*x%mod;
	return res;
}
void init()
{
	phi[1]=1;inv6=power(6,mod-2);
	for(int i=2;i<=N;i++)
	{
		if(!phi[i]) phi[i]=i-1,pri[++tot]=i;
		for(int j=1;j<=tot&&i*pri[j]<=N;j++)
		{
			if(i%pri[j]==0)
			{
				phi[i*pri[j]]=phi[i]*pri[j];
				break;
			}
			phi[i*pri[j]]=phi[i]*(pri[j]-1);
		}
	}
	for(int i=2;i<=N;i++) phi[i]=pls((ll)phi[i]*i%mod*i%mod,phi[i-1]);
}
int work(ll x)
{
	if(x<=N) return phi[(int)x];
	if((itr=ma.find(x))!=ma.end()) return itr->second;
	int res=sum(x);res=(ll)res*res%mod;
	for(ll i=2,j;i<=x;i=j+1)
	{
		j=x/(x/i);
		res=dec(res,(ll)dec(sum2(j),sum2(i-1))*work(x/i)%mod);
	}
	return ma[x]=res;
}
int main()
{
	#ifndef ONLINE_JUDGE
	freopen("in.txt","r",stdin);
	#endif
	read(mod);read(n);
	init();
	for(ll i=1,j,tmp;i<=n;i=j+1)
	{
		j=n/(n/i);tmp=sum(n/i);tmp=(ll)tmp*tmp%mod;
		ans=pls(ans,tmp*dec(work(j),work(i-1))%mod);
	}
	printf("%d\n",ans);
	return 0;
}