题目

题目链接：https://gmoj.net/senior/#main/show/4496
求

\[\sum^{n}_{i=1}\sum_{j|i}\gcd(j,\frac{i}{j}) \]

\(n\leq 10^{11}\)。

思路

\[\sum^{n}_{i=1}\sum_{j|i}\gcd(j,\frac{i}{j}) \]

\[=\sum^{n}_{i=1}\sum^{\lfloor\frac{n}{i}\rfloor}_{j=1}\gcd(i,j) \]

枚举 \(\gcd=d\)

\[\sum^{n}_{d=1}d\sum^{\lfloor\frac{n}{d^2}\rfloor}_{i=1}\sum^{\lfloor\frac{n}{d^2i}\rfloor}_{j=1}[\gcd(i,j)=1] \]

令 \(g(n)=\sum^{n}_{i=1}\sum^{\lfloor\frac{n}{i}\rfloor}_{j=1}[\gcd(i,j)=1]\)，则原式 \(=\sum^{n}_{d=1}g(\lfloor\frac{n}{d^2}\rfloor)\)。

\[g(n)=\sum^{\sqrt{n}}_{d=1}\mu(d)\sum^{\lfloor\frac{n}{d^2}\rfloor}_{i=1}\lfloor\frac{n}{d^2i}\rfloor \]

整除分块即可。记忆化一下可以卡进 \(1s\)。

代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int N=350010,M=10000000;
int m,prm[N],mu[N];
bool v[N];
ll n,ans,f[M];

void findprm(int n)
{
	mu[1]=1;
	for (int i=2;i<=n;i++)
	{
		if (!v[i]) prm[++m]=i,mu[i]=-1;
		for (int j=1;j<=m;j++)
		{
			if (i>n/prm[j]) break;
			v[i*prm[j]]=1; mu[i*prm[j]]=-mu[i];
			if (!(i%prm[j])) { mu[i*prm[j]]=0; break; }
		}
	}
}

ll calc2(ll n)
{
	if (n<M && f[n]) return f[n];
	ll res=0;
	for (ll l=1,r;l<=n;l=r+1)
	{
		r=n/(n/l);
		res+=(r-l+1)*(n/l);
	}
	if (n<M) f[n]=res;
	return res;
}

ll calc1(ll n)
{
	ll res=0;
	for (ll i=1;i*i<=n;i++)
		res+=mu[i]*calc2(n/i/i);
	return res;
}

int main()
{
	freopen("gcd.in","r",stdin);
	freopen("gcd.out","w",stdout);
	findprm(N-1);
	scanf("%lld",&n);
	for (ll i=1;i*i<=n;i++)
		ans+=i*calc1(n/i/i);
	cout<<ans;
	return 0;
}

标签：约数,lfloor,GMOJ4496,frac,gcd,prm,sum,rfloor,互补
来源： https://www.cnblogs.com/stoorz/p/15006209.html