luogu1593 因子和

Description

求 \(\sigma_1(A^B)\mod 9901\)

$1\leqslant A,B\leqslant 50000000 $

Solution

\[A=\prod p_i^{q_i}\ \ \ \ \ A^B=\prod p_i^{q^iB}\\ \sigma_1(A^B)=\sigma_1(\prod p_i^{q^iB})=\prod\sigma_1(p_i^{q_iB})\\ \sigma_1(p^q)=\sum_{i=0}^qp^i \]

然后等比数列求和。

Code

#include<bits/stdc++.h>
using namespace std;
int a,b;
#define MOD 9901
int power(int a,int b)
{
	a %= MOD;
	int res = 1;
	while(b > 0)
	{
		if(b & 1)res = 1ll * res * a % MOD;
		a = 1ll * a * a % MOD;
		b = b >> 1;
	}
	return res;
}
int seq(int a1,int q,long long n)
{
	if(n == 0)return 0;
	if(n == 1)return a1;
	a1 %= MOD;
	if(n % 2 == 0)
	{
		int k = seq(a1,q,n / 2);
		return (k + 1ll * k * power(q,n / 2) % MOD) % MOD;
	}
	else
	{
		return (a1 + seq(1ll * a1 * q % MOD,q,n - 1)) % MOD;
	}
}
int sigma1(int p,long long q)
{//1,p,p^2,...,p^q
	return seq(1,p,q + 1);
}
int fac[100000],cnt[100000],tot = 0;
int main()
{
	cin >> a >> b;
	int k = a;
	for(int i = 2;i <= sqrt(k);++i)
	{
		if(a % i == 0)
		{
			fac[++tot] = i;
			while(a % i == 0)
			{
				++cnt[tot];
				a /= i;
			}
		}
	}
	if(a != 1)
	{
		fac[++tot] = a;
		cnt[tot] = 1;
	}
	//for(int i = 1;i <= tot;++i)cout << fac[i] << " " << cnt[i] << endl;
	int ans = 1;
	for(int i = 1;i <= tot;++i)
	{
		ans = 1ll * ans * sigma1(fac[i],1ll * cnt[i] * b) % MOD;
	}
	cout << ans << endl;
	return 0;
}