题目简介

Description | 题目描述

Consider two natural numbers \(A\) and \(B\). Let \(S\) be the sum of all natural divisors of \(A^B\). Determine \(S\) modulo \(9901\) (the rest of the division of \(S\) by \(9901\)).

给定两个自然数\(A\)和\(B\)，\(S\)为\(A^B\)的所有正整数约数和，编程输出\(S\) \(mod\) \(9901\)的结果。

Input | 输入

The only line contains the two natural numbers \(A\) and \(B\), (\(0\leq A,B \leq 50000000\))separated by blanks.

只有一行，两个用空格隔开的自然数\(A\)和\(B\)（\(0\leq A,B\leq 50000000\)）。

OutPut | 输出

The only line of the output will contain \(S\) modulo \(9901\).

只有一行，即\(S\) \(mod\) \(9901\)的结果。

Sample Input | 样例输入

2 3

Sample Output | 样例输出

15

解法分析

对于自然数\(A\)，一定会有：

\[A = {P_1}^{a_1}\times{P_2}^{a_2}\times……\times{P_n}^{a_n} \]

则：

\[A^B = ({P_1}^{a_1}\times{P_2}^{a_2}\times……\times{P_n}^{a_n})^B \]

\[= {P_1}^{a_1 B}\times{P_2}^{a_2 B}\times……\times{P_n}^{a_n B} \]

其中，\({P_k}^{a_k B}\)的正因数之和为

\[Sum({P_k}^{a_k B})={P_k}^{0}+{P_k}^{1}+{P_k}^{2}+……+{P_k}^{a_k B} \]

所以：

\[S=Sum({P_1}^{a_1 B})\times Sum({P_2}^{a_2 B})\times Sum({P_3}^{a_3 B})\times……\times Sum({P_n}^{a_n}) \]

设 \(b=a_k B\) 如果 \(b\) 为奇数 \((b\in N,b\in O)\)，那么一定有：

\[Sum({P_k}^b)=(1+{P_k}^{\lfloor{\frac{b}{2}}\rfloor}+1)\times Sum({P_k}^{\lfloor{\frac{b}{2}}\rfloor}) \]

可倒推：

\[Sum({P_k}^b)=(1+{P_k}^{\lfloor{\frac{b}{2}}\rfloor+1})\times Sum({P_k}^{\lfloor{\frac{b}{2}}\rfloor}) \]

\[=(1+{P_k}^{\lfloor{\frac{b}{2}}\rfloor+1})\times (1+{P_k}^{1}+{P_k}^{2}+……+{P_k}^{\lfloor{\frac{b}{2}}\rfloor})({P_k}^0=1) \]

\[=1+{P_k}^{1}+{P_k}^{2}+……+{P_k}^{\lfloor{\frac{b}{2}}\rfloor}+{P_k}^{\lfloor{\frac{b}{2}}\rfloor+1}+……+{P_k}^{\lfloor{\frac{b}{2}}\rfloor+\lfloor{\frac{b}{2}}\rfloor+1} \]

由于 \(b\) 为奇数，所以：

\[\lfloor{\frac{b}{2}}\rfloor+\lfloor{\frac{b}{2}}\rfloor+1=b \]

所以：

\[Sum({P_k}^b)={P_k}^{0}+{P_k}^{1}+{P_k}^{2}+……+{P_k}^{b}=(1+{P_k}^{\lfloor{\frac{b}{2}}\rfloor+1})\times sum({P_k}^{\lfloor{\frac{b}{2}}\rfloor)})(b\in N,b\in O) \]

完美成立

如果 \(b\) 为偶数 \((b\in N,b\in E)\)，那么一定有：

\[Sum({P_k}^b)=(1+{P_k}^{\frac{b}{2}})\times Sum({P_k}^{\frac{b}{2}})-{P_k}^{\frac{b}{2}} \]

同理推导：

\[Sum({P_k}^b)=(1+{P_k}^{\frac{b}{2}})\times Sum({P_k}^{\frac{b}{2}})-{P_k}^{\frac{b}{2}} \]

\[=(1+{P_k}^{\frac{b}{2}+1})\times (1+{P_k}^{1}+{P_k}^{2}+……+{P_k}^{\frac{b}{2}})-{P_k}^{\frac{b}{2}} \]

\[={P_k}^{0}+{P_k}^{1}+{P_k}^{2}+……+{P_k}^{\frac{b}{2}}+{P_k}^{\frac{b}{2}}+{P_k}^{\frac{b}{2}+1}……+{P_k}^{b}-{P_k}^{\frac{b}{2}} \]

\[={P_k}^{0}+{P_k}^{1}+{P_k}^{2}+……+{P_k}^{b} \]

所以：

\[Sum({P_k}^b)={P_k}^{0}+{P_k}^{1}+{P_k}^{2}+……+{P_k}^{b}=(1+{P_k}^{\frac{b}{2}})\times Sum({P_k}^{\frac{b}{2}})-{P_k}^{\frac{b}{2}}(b\in N,b\in E) \]

完美成立

而 \(Sum({P_k}^{\frac{b}{2}})\) 又可以由此不断继续分下去
由此，整个问题化解成了 简单的 快速幂+分治

代码

解决这些数学问题后，最后的问题就是long long了
使用 \(p\) 数组存 \(P_k\) , \(a\) 数组存 \(a_k\)

for(int i=2;i<=A;i++){
	if(A%i)continue;
	p[++n]=i;
	while(!(A%i)){
		a[n]++;
		A/=i;
	}
}

快速幂（非递归实现）：

inline ll power(ll a,ll b){
	ll ans=1;
	while(b){
		if(b&1)ans=(ans*a)%mod;
		b>>=1;
		a=(a*a)%mod;
	}
	return ans%mod;
}

分治（很短的）：

inline ll sum(ll c,ll k){
	if(!k)return 1;
	int ret;
	if(k&1)
		ret=(power(c,(k>>1)+1)+1)*sum(c,k>>1)%mod;
	else ret=((power(c,k>>1)+1)*sum(c,k>>1)-power(c,k>>1))%mod;
	return ret%mod;
}

综合起来看就是：

#include<bits/stdc++.h>
typedef long long ll;
const int maxn=10000010;
const int mod=9901;
inline ll power(ll a,ll b){
	ll ans=1;
	while(b){
		if(b&1)ans=(ans*a)%mod;
		b>>=1;
		a=(a*a)%mod;
	}
	return ans%mod;
}
int A,B;
int p[maxn],a[maxn];
int n;
inline ll sum(ll c,ll k){
	if(!k)return 1;
	int ret;
	if(k&1)
		ret=(power(c,(k>>1)+1)+1)*sum(c,k>>1)%mod;
	else ret=((power(c,k>>1)+1)*sum(c,k>>1)-power(c,k>>1))%mod;
	return ret%mod;
}
int main(){
	scanf("%d%d",&A,&B);
	for(int i=2;i<=A;i++){
		if(A%i)continue;
		p[++n]=i;
		while(!(A%i)){
			a[n]++;
			A/=i;
		}
	}
	ll S=1;
	for(int i=1;i<=n;i++){
		S=(S*sum(p[i],a[i]*B))%mod;
	}
	printf("%lld\n",S);
	return 0;
}
$$-----END-----$$