[51Nod 1227] 平均最小公倍数 (杜教筛)

题目描述

求$$\sum _{i=a}^b\sum _{j=1}^i\frac{lcm(i,j)}{i}$$
$1<=a<=b<=10^9$

题目分析

这道题其实是[51Nod 1238] 最小公倍数之和题解的简化版,或者说是本质…
就直接上公式了

令$f(n)={\sum }_{i=1}^n\frac{lcm(n,i)}{n}={\sum }_{i=1}^n\frac{i}{(n,i)}$,则$Ans={\sum }_{i=a}^{b}f(i)$
$$\begin{gathered} f(n)=\sum _{d|n}\sum _{d|i}\frac{i}{d}[(i,n)==d] \\ =\sum _{d|n}\sum _{d|i}\frac{i}{d}[(\frac{i}{d},\frac{n}{d})==1] \\ =\sum _{d|n}\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }i[(i,\frac{n}{d})==1] \\ =\sum _{d|n}\sum _{i=1}^{d}i[(i,d)==1] \end{gathered}$$
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
$$\begin{gathered} \therefore f(n)=\sum _{d|n}\frac{\phi (d)d+[d==1]}{2} \\ =\frac{1+{\sum }_{d|n}\phi (d)d}{2} \end{gathered}$$
$$\begin{gathered} \therefore \sum _{i=1}^{n}f(i)=\frac{n+{\sum }_{i=1}^n{\sum }_{d|n}\phi (d)d}{2} \\ = \\ \frac{n+{\sum }_{d=1}^n\phi (d)d{\sum }_{d|i}1}{2} \\ =\frac{n+{\sum }_{d=1}^n\phi (d)d\lfloor \frac{n}{d}\rfloor }{2} \end{gathered}$$
此时就可以用整除分块优化+杜教筛计算${\sum }_{d=1}^n\phi (d)d$
令$h(n)=\phi (d)d,g(n)={\sum }_{i=1}^{n}h(d)$
$$\begin{gathered} \because n=\sum _{d|n}\phi (d) \\ \therefore n^2=\sum _{d|n}\phi (d)n=\sum _{d|n}\phi (d)d\cdot \frac{n}{d}=\sum _{d|n}h(d)\frac{n}{d} \\ \therefore \sum _{i=1}^n{i}^2=\sum _{i=1}^n\sum _{d|i}h(d)\frac{i}{d} \\ =\sum _{d=1}^{n}h(d)\sum _{d|i}\frac{i}{d} \\ =\sum _{d=1}^{n}h(d)\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }i \end{gathered}$$
注意我们是在杜教筛,不能到这里就把${\sum }_{i=1}^{\lfloor \frac{n}{d}\rfloor }i$看做$\Theta (1)$可求的式子而之后再也不做变换,那样往往会陷入更麻烦的方法或者死胡同里去,接着往下
$$\begin{gathered} \therefore \sum _{i=1}^n{i}^2=\sum _{i=1}^{n}i\sum _{d=1}^{\lfloor \frac{n}{i}\rfloor }h(d)=\sum _{i=1}^{n}i\cdot g(\lfloor \frac{n}{i}\rfloor ) \\ \therefore g(n)=\sum _{i=1}^n{i}^2-\sum _{i=2}^{n}i\cdot g(\lfloor \frac{n}{i}\rfloor ) \end{gathered}$$
然后套杜教筛即可

虽然在外面套了一层整除分块优化,但由于记忆化的原因,不影响时间复杂度,预处理出一部分$g$后总复杂度为$\Theta (n^{\frac{2}{3}})$

…有兴趣的可以去了解一下[51Nod 1238] 最小公倍数之和题解,比这道题恶心点

AC code

#include 
#include 
#include 
#include 
#include 
using namespace std;
using namespace tr1;
typedef long long LL;
const int inv6 = 166666668;
const int inv2 = 500000004;
const int MAXN = 5e6 + 1;
const int mod = 1e9 + 7;
int Prime[MAXN], phi[MAXN], Cnt;
bool IsnotPrime[MAXN];
LL g[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) % mod;
}
unordered_map<LL, LL> G;
inline LL sum1(LL i, LL j) { return ((i+j)%mod) * ((j-i+1)%mod) % mod * inv2 % mod; }
inline LL sum(LL n)
{
	if(n < MAXN) return g[n];
	if(G.count(n)) return G[n];
	LL ret = (n%mod) * ((n+1)%mod) % mod * ((2*n+1)%mod) % mod * inv6 % mod;
	for(LL i = 2, j; i <= n; i=j+1)
	{
		j = n/(n/i);
		ret = (ret - sum(n/i) * sum1(i, j) % mod) % mod;
	}
	return G[n]=ret;
}

LL solve(LL n)
{
	LL ret = n%mod;
	for(LL i = 1, j; i <= n; i=j+1)
	{
		j = n/(n/i);
		ret = (ret + ((sum(j)-sum(i-1))%mod) * ((n/i)%mod) % mod) % mod;
	}
	return ret * inv2 % mod;
}

int main ()
{
	LL a, b; init();
	scanf("%lld%lld", &a, &b);
	printf("%lld\n", ((solve(b)-solve(a-1)) % mod + mod) % mod);
}