Luogu P1593 因子和

题目

求\(A^B\)的因子和对\(9901\)取模后的结果

题解

对\(A\)分解质因数\(A=\sum p_{i}^{c_i}\)
那么\(A^B = \sum p_{i}^{c_i\times B}\)
考虑\(A^B\)的所有因数和可表示为\((1+p_{1}+p_{1}^{2}+...+p_{1}^{c_1 \times B}) \times (1+p_{2}+p_{2}^{2}+...+p_{2}^{c_2 \times B}) \times ... \times (1+p_{n}+p_{n}^{2}+...+p_{n}^{c_n \times B})\)
最后结果为若干等比数列和的乘积，但是等比数列求和涉及到除法，除法的取余运算涉及到逆元的知识，这里我们换个角度思考。
定义\(sum(p,c)=1+p+p^2+...+p^c\)
当\(c\)为奇数时\(sum(p, c) = 1+p+p^2+...+p^{\frac{c-1}{2}}+p^{\frac{c+1}{2}}+...+p^c=(1+p+...+p^{\frac{c-1}{2}})+p^{\frac{c+1}{2}} \times (1+p+...+p^{\frac{c-1}{2}})=(1+p^{\frac{c+1}{2}})\times sum(p, \frac{c-1}{2})\)
同理\(c\)为偶数时\(sum(p, c) = (1 + p^{\frac{c}{2}}) \times sum(p, \frac{c}{2} - 1) + p ^ c\)
分治求解即可

code

#include <bits/stdc++.h> 

using namespace std; 
typedef long long LL; 

const LL mod = 9901; 

LL quickm(LL a, LL b) {
	LL ans = 1ll; 
	for (; b; b >>= 1) {
		if (b & 1) ans = (LL)(ans * a) % mod; 
		a = (LL)(a * a) % mod; 
	}
	return ans % mod; 
}

// x^y
// y为奇数 (1 + x^((y+1)/2)) * sum(x, (y-1)/2)
// y为偶数 (1 + x^(y/2)) * sum(x, y/2 - 1) + x ^ y
LL sum(LL x, LL y) {
	if (y == 0) return 1; 
	if (y % 2 == 1) {
		return (LL)(1 + quickm(x, (y+1)/2)) * sum(x, (y-1)/2) % mod; 
	} else {
		return (LL)((1 + quickm(x, y/2)) * sum(x, (y/2)-1) + quickm(x, y)) % mod; 
	}
}
 
int main() {
	int a, b; 
	scanf("%d%d", &a, &b); 
	if (a == 0) {
		cout << 0 << endl; 
		return 0; 
	}

	// 分解质因子
	LL temp = a, res = 1; 
	for (int i =  2; i * i <= a; ++i) {
		if (temp % i == 0) {
			LL x = i, y = 0;
			for (; temp % i == 0; temp /= i) ++y; 
	 		res = (res * sum(x, (LL)y * b)) % mod; 
		}
	}
	if (temp > 1) 
		res = res * sum(temp, b) % mod; 
 	printf("%lld\n", res);
	return 0; 
}