[51Nod 1237] 最大公约数之和 (杜教筛+莫比乌斯反演)

题目描述

求$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^n(i,j)mod(1e9+7) \\ n<=10^{10} \end{gathered}$$

题目分析

乍一看十分像裸莫比乌斯反演,然而$n$的范围让人望而却步
于是先变化一下式子
$$Ans=\sum _{i=1}^n\sum _{j=1}^n(i,j)$$
枚举$T=(i,j)$
$$\begin{gathered} =\sum _{T=1}^n\sum _{i=1}^{\lfloor \frac{n}{T}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{T}\rfloor }[(i,j)==1] \\ =\sum _{T=1}^n\sum _{i=1}^{\lfloor \frac{n}{T}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{T}\rfloor }\sum _{d|i,d|j}\mu (d) \\ =\sum _{T=1}^{n}T\sum _{d=1}^{\lfloor \frac{n}{T}\rfloor }\mu (d){\lfloor \frac{n}{Td}\rfloor }^2 \end{gathered}$$
令k=Td
$$=\sum _{k=1}^n{\lfloor \frac{n}{k}\rfloor }^2\phi (k)$$
则此时可以整除分块优化,每次算出$\lfloor \frac{n}{k}\rfloor$相等的上下界$i,j$后用莫比乌斯反演计算$(S\phi (j)-S\phi (i-1))$
由于计算$\phi$的前缀和时记忆化处理过,所以在杜教筛外面再套了一个整除分块优化不会影响时间复杂度,复杂度仍是$\Theta (n^{\frac{2}{3}})$

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;
map<LL, LL> S; LL s[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) s[i] = (s[i-1] + phi[i]) % mod;
}

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

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