【BZOJ4805】欧拉函数求和（杜教筛）

【BZOJ4805】欧拉函数求和（杜教筛）

题面

BZOJ

题解

好久没写过了

正好看见了顺手切一下

令$$S(n)=\sum_{i=1}^n\varphi(i)$$

设存在的某个积性函数\(g(x)\)

\[(g*\varphi)(i)=\sum_{d|i}g(d)\varphi(\frac{i}{d})
\]

\[\sum_{i=1}^n(g*\varphi(i))(i)
\]

\[=\sum_{i=1}^n\sum_{d|i}g(d)\varphi(\frac{i}{d})
\]

\[=\sum_{d=1}^ng(d)\sum_{d|i}\varphi(\frac{i}{d})
\]

\[=\sum_{d=1}^ng(d)\sum_{i=1}^{n/i}\varphi(i)
\]

\[=\sum_{d=1}^ng(d)S(\frac{n}{d})
\]

拿出杜教筛的套路柿子

\[g(1)S(n)=\sum_{i=1}^n(g*\varphi)(i)-\sum_{i=2}^ng(i)S(\frac{n}{i})
\]

我们知道\((\varphi*1)=x\)

\[S(n)=\sum_{i=1}^ni-\sum_{i=2}^nS(\frac{n}{i})
\]

\[S(n)=\frac{n(n+1)}{2}-\sum_{i=2}^nS(\frac{n}{i})
\]

预处理\(10^7\)然后杜教筛美滋滋

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<queue>
using namespace std;
#define ll long long
#define RG register
#define MAX 10000000
inline int read()
{
    RG int x=0,t=1;RG char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
    if(ch=='-')t=-1,ch=getchar();
    while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
    return x*t;
}
int pri[MAX+10],tot;
bool zs[MAX+10];
ll phi[MAX+10];
void pre()
{
	zs[1]=true;phi[1]=1;
	for(int i=2;i<=MAX;++i)
	{
		if(!zs[i])pri[++tot]=i,phi[i]=i-1;
		for(int j=1;j<=tot&&pri[j]*i<=MAX;++j)
		{
			zs[i*pri[j]]=true;
			if(i%pri[j])phi[i*pri[j]]=phi[i]*phi[pri[j]];
			else{phi[i*pri[j]]=phi[i]*pri[j];break;}
		}
	}
	for(int i=1;i<=MAX;++i)phi[i]+=phi[i-1];
}
map<ll,ll> M;
ll Solve(ll x)
{
	if(x<=MAX)return phi[x];
	if(M[x])return M[x];
	ll ret=0;
	for(ll i=2,j;i<=x;i=j+1)
	{
		j=x/(x/i);
		ret+=(j-i+1)*Solve(x/i);
	}
	return M[x]=x*(x+1)/2-ret;
}
int main()
{
	pre();
	int n=read();
	printf("%lld\n",Solve(n));
	return 0;
}