poj 2480 Longge's problem 积性函数性质+欧拉函数

题意:

求f(n)=∑gcd(i, N) 1<=i <=N.

分析:

f(n)是积性的数论上有证明(f(n)=sigma{1<=i<=N} gcd(i,N) = sigma{d | n}phi(n / d) * d ,后者是积性函数),可以这么解释:当d是n的因子时,设1至n内有a1,a2,..ak满足gcd(n,ai)==d,那么d这个因子贡献是d*k,接下来证明k=phi(n/d):设gcd(x,n)==d,那么gcd(x/d,n/d)==1,所以满足条件的x/d数目为phi(n/d),x的数目也为phi(n/d)。

代码:

//poj 2480
//sep9
/*
f(pi^ai) =  Φ(pi^ai)+pi*Φ(pi^(ai-1))+pi^2*Φ(pi^(ai-2))+...+pi^(ai-1)* Φ(pi)+ pi^ai *Φ(1)
     = pi^(ai-1)*(pi-1) + pi*pi^(ai-2)*(pi-1)....+pi^ai
     =  pi^ai*(1+ai*(1-1/pi))
f(n) = p1^a1*p2^a2...*pr^ar*(1+a1*(1-1/p1))*(1+a2*(1-1/p2))*...
       =  n*(1+a1*(1-1/p1))*(1+a2*(1-1/p2))*...

*/
#include <iostream>
using namespace std;
typedef long long ll;

int main()
{
	ll n;
	while(scanf("%lld",&n)==1){
		ll ans=n;
		for(ll i=2;i*i<=n;++i){
			if(n%i==0){
				ll a=0,p=i;
				while(n%p==0){
					++a;
					n/=p;
				}
				ans=ans+ans*a*(p-1)/p;
			}
		}
<span style="white-space:pre">		</span>if(n!=1)
<span style="white-space:pre">			</span>ans=ans+ans*(n-1)/n;
		printf("%I64d\n",ans);
	}
	return 0;
}
 

poj 2480 Longge's problem 积性函数性质+欧拉函数

时间: 2024-10-15 08:54:34

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