[51Nod 1237] 最大公约数之和 (杜教筛+莫比乌斯反演)

题目描述

求∑i=1n∑j=1n(i,j) mod (1e9+7)n&lt;=1010\sum_{i=1}^n\sum_{j=1}^n(i,j)~mod~(1e9+7)\\n&lt;=10^{10}i=1∑n​j=1∑n​(i,j) mod (1e9+7)n<=1010

题目分析

乍一看十分像裸莫比乌斯反演,然而nnn的范围让人望而却步

于是先变化一下式子

Ans=∑i=1n∑j=1n(i,j)Ans=\sum_{i=1}^n\sum_{j=1}^n(i,j)Ans=i=1∑n​j=1∑n​(i,j)

枚举T=(i,j)T=(i,j)T=(i,j)

=∑T=1n∑i=1⌊nT⌋∑j=1⌊nT⌋[(i,j)==1]=∑T=1n∑i=1⌊nT⌋∑j=1⌊nT⌋∑d∣i,d∣jμ(d)=∑T=1nT∑d=1⌊nT⌋μ(d)⌊nTd⌋2=\sum_{T=1}^n\sum_{i=1}^{\lfloor\frac nT\rfloor}\sum_{j=1}^{\lfloor\frac nT\rfloor}[(i,j)==1]\\=\sum_{T=1}^n\sum_{i=1}^{\lfloor\frac nT\rfloor}\sum_{j=1}^{\lfloor\frac nT\rfloor}\sum_{d|i,d|j}\mu(d)\\=\sum_{T=1}^nT\sum_{d=1}^{\lfloor\frac nT\rfloor}\mu(d){\lfloor\frac n{Td}\rfloor}^2=T=1∑n​i=1∑⌊Tn​⌋​j=1∑⌊Tn​⌋​[(i,j)==1]=T=1∑n​i=1∑⌊Tn​⌋​j=1∑⌊Tn​⌋​d∣i,d∣j∑​μ(d)=T=1∑n​Td=1∑⌊Tn​⌋​μ(d)⌊Tdn​⌋2

令k=Td

=∑k=1n⌊nk⌋2φ(k)=\sum_{k=1}^n{\lfloor\frac n{k}\rfloor}^2\varphi(k)=k=1∑n​⌊kn​⌋2φ(k)

则此时可以整除分块优化,每次算出⌊nk⌋{\lfloor\frac n{k}\rfloor}⌊kn​⌋相等的上下界i,ji,ji,j后用莫比乌斯反演计算(Sφ(j)−Sφ(i−1))(S\varphi(j)-S\varphi(i-1))(Sφ(j)−Sφ(i−1))

由于计算φ\varphiφ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Θ(n23)\Theta(n^{\frac23})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;
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);
}