POJ 2480 Longge's problem 積性函數

POJ 2480 Longge's problem 積性函數

題目來源：POJ 2480 Longge's problem

題意：求i從1到n的gcd(n, i)的和

思路：首先如果m, n 互質 gcd(i, n*m) = gcd(i, n)*gcd(i, m) 這是一個積性函數積性函數的和還是積性函數

由歐拉函數知識得 phi(p^a) = p^a - p^(a-1) p是素數 a是正整數

得到最終答案f(n) = f(p1^a1)*f(p2^a2)*...*f(pn^an) 其中f(p^a) = a*(p^a-p^(a-1))+p^a

#include 
#include 
#include 
#include 
using namespace std;
typedef long long LL;

const int maxn = 1000010;
//篩素數 
int vis[maxn];
LL prime[maxn];

LL pow_mod(LL a, LL p)
{
	LL ans = 1;
	while(p)
	{
		if(p&1)
		{
			ans *= a;

		}
		a *= a;

		p >>= 1;
	}
	return ans;
}

void sieve(int n)
{
	int m = sqrt(n+0.5);
	memset(vis, 0, sizeof(vis));
	vis[0] = vis[1] = 1;
	for(int i = 2; i <= m; i++)
		if(!vis[i])
			for(int j = i*i; j <= n; j += i)
				vis[j] = 1;
}

int get_primes(int n)
{
	sieve(n);
	int c = 0;
	for(int i = 2; i <= n; i++)
		if(!vis[i])
			prime[c++] = i;
	return c;
}

int main()
{
	int c = get_primes(200000);
	int cas = 1;
	int T;
	LL n, ans;
	while(scanf("%I64d", &n) != EOF)
	{
		ans = 1;	
		for(int i = 0; i < c && prime[i]*prime[i] <= n; i++)
		{
			if(n%prime[i] == 0)
			{
				LL sum = 0;
				while(n%prime[i] == 0)
				{
					sum++;
					n /= prime[i];
					
				}
				ans *= sum*(pow_mod(prime[i], sum)-pow_mod(prime[i], sum-1))+pow_mod(prime[i], sum);
			}
		}
		if(n > 1)
		{
			ans *= n-1+n;
		}
		printf("%I64d\n", ans);
	}
	return 0;
}