[51Nod 1238] 最小公倍数之和 (恶心杜教筛)

题目描述

求$sum_{i=1}^Nsum_{j=1}^Nlcm(i,j)$
$2<=N<=10^{10}$

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
$large Ans=sum_{i=1}^nsum_{j=1}^nlcm(i,j)=2sum_{i=1}^nsum_{j=1}^ilcm(i,j)-frac{n(n+1)}{2}\Let~s(n)=sum_{i=1}^nsum_{j=1}^ilcm(i,j),f(n)=sum_{i=1}^nlcm(i,n)\ herefore f(n)=sum_{i=1}^nfrac{in}{(i,n)}=nsum_{i=1}^nfrac i{(i,n)}\=nsum_{d|n}sum_{i=1}^n[(i,n)==d]frac id\=nsum_{d|n}sum_{i=1}^{frac nd}[(i,frac nd)==1]i\=nsum_{d|n}sum_{i=1}^d[(i,d)==1]i\=nsum_{d|n}frac{varphi(d)d+[d==1]}2$
此处有一个常识
$sum_{i=1}^ni[(i,n)==1]=frac {varphi(n)n+[n==1]}2$

• 证明如下
  • 当$n>1$时,若$(i,n)=1iff(n-i,n)=1$,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
    $large sum_{i=1}^ni[(i,n)==1]=frac {varphi(n)n+[n==1]}2$

继续
$large herefore f(n)=ncdotfrac {1+sum_{d|n}varphi(d)d}2\s(n)=sum_{i=1}^nf(i)=frac{sum_{i=1}^ni(1+sum_{d|i}varphi(d)d)}2\=frac{frac{n(n+1)}2+sum_{i=1}^nisum_{d|i}varphi(d)d}2\=frac{frac{n(n+1)}2+sum_{d=1}^nvarphi(d)dsum_{d|i}i}2\=frac{frac{n(n+1)}2+sum_{d=1}^nvarphi(d)d^2sum_{i=1}^{lfloorfrac nd floor}i}2\=frac{frac{n(n+1)}2+sum_{i=1}^nisum_{d=1}^{lfloorfrac ni floor}varphi(d)d^2}2\Ans=2s(n)-frac{n(n+1)}2=sum_{i=1}^nisum_{d=1}^{lfloorfrac ni floor}varphi(d)d^2\Let~h(d)=varphi(d)d^2,g(n)=sum_{d=1}^nh(d)\n=sum_{d|n}varphi(d)\n^3=sum_{d|n}varphi(d)n^2\=sum_{d|n}varphi(d)d^2(frac nd)^2\=sum_{d|n}h(d)(frac nd)^2\sum_{i=1}^ni^3=sum_{i=1}^nsum_{d|n}h(d)(frac id)^2\=sum_{d=1}^nh(d)sum_{d|i}(frac id)^2\=sum_{d=1}^nh(d)sum_{i=1}^{lfloorfrac nd floor}i^2\=sum_{i=1}^ni^2sum_{d=1}^{lfloorfrac ni floor}h(d)\=sum_{i=1}^ni^2g(lfloorfrac ni floor)\g(n)=sum_{i=1}^ni^3-sum_{i=2}^ni^2g(lfloorfrac ni floor)\=(frac{n(n+1)}2)^2-sum_{i=2}^ni^2g(lfloorfrac ni floor)$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$large Ans=sum_{i=1}^nicdot g(lfloorfrac ni floor)$

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$Theta(n^{frac 23})$

AC code

#include <cstdio>
#include <map>
#include <cstring>
#include <algorithm>
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
", (solve(n)+mod)%mod);
}