[51Nod 1238] 最小公倍数之和 (恶心杜教筛)

题目描述

求∑i=1N∑j=1Nlcm(i,j)\sum_{i=1}^N\sum_{j=1}^Nlcm(i,j)i=1∑N​j=1∑N​lcm(i,j)

2&lt;=N&lt;=10102&lt;=N&lt;=10^{10}2<=N<=1010

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101010^{10}1010,莫比乌斯反演不行了了,所以我们看看怎样玄学杜教筛

Ans=∑i=1n∑j=1nlcm(i,j)=2∑i=1n∑j=1ilcm(i,j)−n(n+1)2Let s(n)=∑i=1n∑j=1ilcm(i,j),f(n)=∑i=1nlcm(i,n)∴f(n)=∑i=1nin(i,n)=n∑i=1ni(i,n)=n∑d∣n∑i=1n[(i,n)==d]id=n∑d∣n∑i=1nd[(i,nd)==1]i=n∑d∣n∑i=1d[(i,d)==1]i=n∑d∣nφ(d)d+[d==1]2\large Ans=\sum_{i=1}^n\sum_{j=1}^nlcm(i,j)=2\sum_{i=1}^n\sum_{j=1}^ilcm(i,j)-\frac{n(n+1)}{2}\\Let~s(n)=\sum_{i=1}^n\sum_{j=1}^ilcm(i,j),f(n)=\sum_{i=1}^nlcm(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 id\\=n\sum_{d|n}\sum_{i=1}^{\frac nd}[(i,\frac nd)==1]i\\=n\sum_{d|n}\sum_{i=1}^d[(i,d)==1]i\\=n\sum_{d|n}\frac{\varphi(d)d+[d==1]}2Ans=i=1∑n​j=1∑n​lcm(i,j)=2i=1∑n​j=1∑i​lcm(i,j)−2n(n+1)​Let s(n)=i=1∑n​j=1∑i​lcm(i,j),f(n)=i=1∑n​lcm(i,n)∴f(n)=i=1∑n​(i,n)in​=ni=1∑n​(i,n)i​=nd∣n∑​i=1∑n​[(i,n)==d]di​=nd∣n∑​i=1∑dn​​[(i,dn​)==1]i=nd∣n∑​i=1∑d​[(i,d)==1]i=nd∣n∑​2φ(d)d+[d==1]​

此处有一个常识

∑i=1ni[(i,n)==1]=φ(n)n+[n==1]2\sum_{i=1}^ni[(i,n)==1]=\frac {\varphi(n)n+[n==1]}2i=1∑n​i[(i,n)==1]=2φ(n)n+[n==1]​

• 证明如下
  • 当n&gt;1n&gt;1n>1时,若(i,n)=1&ThickSpace;⟺&ThickSpace;(n−i,n)=1(i,n)=1\iff(n-i,n)=1(i,n)=1⟺(n−i,n)=1,所以与nnn互质的数是成对出现,且他们的和为nnn
  • 再加之n=1n=1n=1的特殊情况,可得
    
    ∑i=1ni[(i,n)==1]=φ(n)n+[n==1]2\large \sum_{i=1}^ni[(i,n)==1]=\frac {\varphi(n)n+[n==1]}2∑i=1n​i[(i,n)==1]=2φ(n)n+[n==1]​

继续

∴f(n)=n⋅1+∑d∣nφ(d)d2s(n)=∑i=1nf(i)=∑i=1ni(1+∑d∣iφ(d)d)2=n(n+1)2+∑i=1ni∑d∣iφ(d)d2=n(n+1)2+∑d=1nφ(d)d∑d∣ii2=n(n+1)2+∑d=1nφ(d)d2∑i=1⌊nd⌋i2=n(n+1)2+∑i=1ni∑d=1⌊ni⌋φ(d)d22Ans=2s(n)−n(n+1)2=∑i=1ni∑d=1⌊ni⌋φ(d)d2Let h(d)=φ(d)d2,g(n)=∑d=1nh(d)n=∑d∣nφ(d)n3=∑d∣nφ(d)n2=∑d∣nφ(d)d2(nd)2=∑d∣nh(d)(nd)2∑i=1ni3=∑i=1n∑d∣nh(d)(id)2=∑d=1nh(d)∑d∣i(id)2=∑d=1nh(d)∑i=1⌊nd⌋i2=∑i=1ni2∑d=1⌊ni⌋h(d)=∑i=1ni2g(⌊ni⌋)g(n)=∑i=1ni3−∑i=2ni2g(⌊ni⌋)=(n(n+1)2)2−∑i=2ni2g(⌊ni⌋)
\large \therefore f(n)=n\cdot\frac {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}^ni\sum_{d|i}\varphi(d)d}2\\=\frac{\frac{n(n+1)}2+\sum_{d=1}^n\varphi(d)d\sum_{d|i}i}2\\=\frac{\frac{n(n+1)}2+\sum_{d=1}^n\varphi(d)d^2\sum_{i=1}^{\lfloor\frac nd\rfloor}i}2\\=\frac{\frac{n(n+1)}2+\sum_{i=1}^ni\sum_{d=1}^{\lfloor\frac ni\rfloor}\varphi(d)d^2}2\\Ans=2s(n)-\frac{n(n+1)}2=\sum_{i=1}^ni\sum_{d=1}^{\lfloor\frac ni\rfloor}\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}^n\sum_{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}^{\lfloor\frac nd\rfloor}i^2\\=\sum_{i=1}^ni^2\sum_{d=1}^{\lfloor\frac ni\rfloor}h(d)\\=\sum_{i=1}^ni^2g(\lfloor\frac ni\rfloor)\\g(n)=\sum_{i=1}^ni^3-\sum_{i=2}^ni^2g(\lfloor\frac ni\rfloor)\\=(\frac{n(n+1)}2)^2-\sum_{i=2}^ni^2g(\lfloor\frac ni\rfloor)∴f(n)=n⋅21+∑d∣n​φ(d)d​s(n)=i=1∑n​f(i)=2∑i=1n​i(1+∑d∣i​φ(d)d)​=22n(n+1)​+∑i=1n​i∑d∣i​φ(d)d​=22n(n+1)​+∑d=1n​φ(d)d∑d∣i​i​=22n(n+1)​+∑d=1n​φ(d)d2∑i=1⌊dn​⌋​i​=22n(n+1)​+∑i=1n​i∑d=1⌊in​⌋​φ(d)d2​Ans=2s(n)−2n(n+1)​=i=1∑n​id=1∑⌊in​⌋​φ(d)d2Let h(d)=φ(d)d2,g(n)=d=1∑n​h(d)n=d∣n∑​φ(d)n3=d∣n∑​φ(d)n2=d∣n∑​φ(d)d2(dn​)2=d∣n∑​h(d)(dn​)2i=1∑n​i3=i=1∑n​d∣n∑​h(d)(di​)2=d=1∑n​h(d)d∣i∑​(di​)2=d=1∑n​h(d)i=1∑⌊dn​⌋​i2=i=1∑n​i2d=1∑⌊in​⌋​h(d)=i=1∑n​i2g(⌊in​⌋)g(n)=i=1∑n​i3−i=2∑n​i2g(⌊in​⌋)=(2n(n+1)​)2−i=2∑n​i2g(⌊in​⌋)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ggg然后较大的就用杜教筛计算

又Ans=∑i=1ni⋅g(⌊ni⌋)\large Ans=\sum_{i=1}^ni\cdot g(\lfloor\frac ni\rfloor)Ans=∑i=1n​i⋅g(⌊in​⌋)

因为ggg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ggg函数的时间复杂度,所以复杂度为Θ(n23)\Theta(n^{\frac 23})Θ(n32​)

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\n", (solve(n)+mod)%mod);
}