[51Nod 1238] 最小公倍数之和 (恶心杜教筛)

题目描述

求$$\sum _{i=1}^N\sum _{j=1}^{N}lcm(i,j)$$
$2<=N<=10^{10}$

题目分析

这道题题面跟[bzoj 2693] jzptab & [bzoj 2154] Crash的数字表格 一样,然而数据范围加强到了$10^{10}$,莫比乌斯反演不行了了,所以我们看看怎样玄学杜教筛
$$\begin{gathered} Ans=\sum _{i=1}^n\sum _{j=1}^{n}lcm(i,j)=2\sum _{i=1}^n\sum _{j=1}^{i}lcm(i,j)-\frac{n(n+1)}{2} \\ Lets(n)=\sum _{i=1}^n\sum _{j=1}^{i}lcm(i,j),f(n)=\sum _{i=1}^{n}lcm(i,n) \\ \therefore f(n)=\sum _{i=1}^n\frac{in}{(i,n)}=n\sum _{i=1}^n\frac{i}{(i,n)} \\ =n\sum _{d|n}\sum _{i=1}^n[(i,n)==d]\frac{i}{d} \\ =n\sum _{d|n}\sum _{i=1}^{\frac{n}{d}}[(i,\frac{n}{d})==1]i \\ =n\sum _{d|n}\sum _{i=1}^d[(i,d)==1]i \\ =n\sum _{d|n}\frac{\phi (d)d+[d==1]}{2} \end{gathered}$$
此处有一个常识
$$\sum _{i=1}^{n}i[(i,n)==1]=\frac{\phi (n)n+[n==1]}{2}$$

• 证明如下
  • 当$n>1$时,若$(i,n)=1⟺(n-i,n)=1$,所以与$n$互质的数是成对出现,且他们的和为$n$
  • 再加之$n=1$的特殊情况,可得
    ${\sum }_{i=1}^{n}i[(i,n)==1]=\frac{\phi (n)n+[n==1]}{2}$

继续
$$\begin{gathered} \therefore f(n)=n\cdot \frac{1+{\sum }_{d|n}\phi (d)d}{2} \\ s(n)=\sum _{i=1}^{n}f(i)=\frac{{\sum }_{i=1}^{n}i(1+{\sum }_{d|i}\phi (d)d)}{2} \\ =\frac{\frac{n(n+1)}{2}+{\sum }_{i=1}^{n}i{\sum }_{d|i}\phi (d)d}{2} \\ =\frac{\frac{n(n+1)}{2}+{\sum }_{d=1}^n\phi (d)d{\sum }_{d|i}i}{2} \\ =\frac{\frac{n(n+1)}{2}+{\sum }_{d=1}^n\phi (d)d^2{\sum }_{i=1}^{\lfloor \frac{n}{d}\rfloor }i}{2} \\ =\frac{\frac{n(n+1)}{2}+{\sum }_{i=1}^{n}i{\sum }_{d=1}^{\lfloor \frac{n}{i}\rfloor }\phi (d)d^2}{2} \\ Ans=2s(n)-\frac{n(n+1)}{2}=\sum _{i=1}^{n}i\sum _{d=1}^{\lfloor \frac{n}{i}\rfloor }\phi (d)d^2 \\ Leth(d)=\phi (d)d^2,g(n)=\sum _{d=1}^{n}h(d) \\ n=\sum _{d|n}\phi (d) \\ {n}^3=\sum _{d|n}\phi (d)n^2 \\ =\sum _{d|n}\phi (d)d^2(\frac{n}{d}{)}^2 \\ =\sum _{d|n}h(d)(\frac{n}{d}{)}^2 \\ \sum _{i=1}^n{i}^3=\sum _{i=1}^n\sum _{d|i}h(d)(\frac{i}{d}{)}^2 \\ =\sum _{d=1}^{n}h(d)\sum _{d|i}(\frac{i}{d}{)}^2 \\ =\sum _{d=1}^{n}h(d)\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }{i}^2 \\ =\sum _{i=1}^n{i}^2\sum _{d=1}^{\lfloor \frac{n}{i}\rfloor }h(d) \\ =\sum _{i=1}^n{i}^{2}g(\lfloor \frac{n}{i}\rfloor ) \\ g(n)=\sum _{i=1}^n{i}^3-\sum _{i=2}^n{i}^{2}g(\lfloor \frac{n}{i}\rfloor ) \\ =(\frac{n(n+1)}{2}{)}^2-\sum _{i=2}^n{i}^{2}g(\lfloor \frac{n}{i}\rfloor ) \end{gathered}$$然后就是杜教筛的形式了,上杜教筛即可.先预处理出小范围的$g$然后较大的就用杜教筛计算

又$Ans={\sum }_{i=1}^{n}i\cdot g(\lfloor \frac{n}{i}\rfloor )$

因为$g$函数求解时是用的记忆化,所以在外面套上一层分块优化不会影响$g$函数的时间复杂度,所以复杂度为$\Theta (n^{\frac{2}{3}})$

AC code

#include 
#include 
#include 
#include 
using namespace std;
typedef long long LL;
const int mod = 1e9+7;
const int MAXN = 5e6+1;
const int inv2 = 500000004;
const int inv3 = 333333336;
map<LL, LL> G; LL g[MAXN];
int Prime[MAXN], Cnt, phi[MAXN];
bool IsnotPrime[MAXN];

void init()
{
	phi[1] = 1;
	for(int i = 2; i < MAXN; ++i)
	{
		if(!IsnotPrime[i]) Prime[++Cnt] = i, phi[i] = i-1;
		for(int j = 1; j <= Cnt && i * Prime[j] < MAXN; ++j)
		{
			IsnotPrime[i * Prime[j]] = 1;
			if(i % Prime[j] == 0)
			{
				phi[i * Prime[j]] = phi[i] * Prime[j];
				break;
			}
			phi[i * Prime[j]] = phi[i] * phi[Prime[j]];
		}
	}
	for(int i = 1; i < MAXN; ++i) g[i] = (g[i-1] + 1ll * phi[i] * i % mod * i % mod) % mod;
}

inline LL sum2(LL i) { return (i%mod) * ((i+1)%mod) % mod * ((2*i+1)%mod) % mod * inv2 % mod * inv3 % mod; }

inline LL calc(LL n)
{
	if(n < MAXN) return g[n];
	if(G.count(n)) return G[n];
	LL ret = (n%mod) * ((n+1)%mod) % mod * inv2 % mod;
	ret = ret * ret % mod;
	for(LL i = 2, j; i <= n; i=j+1)
	{
		j = n/(n/i);
		ret = (ret - (sum2(j)-sum2(i-1)) % mod * calc(n/i) % mod) % mod;
	}
	return G[n] = ret;
}

inline LL sum(LL i, LL j) { return ((i+j)%mod) * ((j-i+1)%mod) % mod * inv2 % mod; }

inline LL solve(LL n)
{
	LL ret = 0;
	for(LL i = 1, j; i <= n; i=j+1)
	{
		j = n/(n/i);
		ret = (ret + sum(i, j) * calc(n/i) % mod) % mod;
	}
	return ret;
}
int main ()
{
	init(); LL n;
	scanf("%lld", &n);
	printf("%lld\n", (solve(n)+mod)%mod);
}