题意：求(A^B)的所有约数之和$ mod 9901(1<=A,B<=5*10^7)$
分析：把A分解质因数即(A=p_1^{c_1}*p_2^{c_2}*...*p_m^{c_m}),则(A^B=p_1^{B*c_1}*p_2^{B*c_2}*...*p_m^{B*c_m}),则(A^B)的所有约数之和就是((1+p_1+p_1^2+...+p_1^{B*c_1})*(1+p_2+p_2^2+...+p_2^{B*c_2})*...*(1+p_m+p_m^2+...+p_m^{B*c_m})).然后对于每一组用分治法来求即可.
具体来说，要求(sum(p,c)=a+p+p^2+...+p^c)?若c为奇数，(sum(p,c)=(1+p^{frac{c+1}{2}})*sum(p,frac{c-1}{2}))；若c为偶数，(sum(p,c)=(1+p^{frac{c}{2}})*sum(p,frac{c}{2}-1)+p^c).
注意有些乘的地方要long long,然后不知道为什么竟然不需要特判...
//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define LL long long
using namespace std;
inline int read() {
int x=0,o=1;char ch=getchar();
while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
if(ch=='-')o=-1,ch=getchar();
while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
return x*o;
}
const int mod=9901;
int A,B,m,p[100005],c[100005];
inline void Divide(int n){
m=0;
/*if(n==2||n==3){
m=1;p[m]=n;c[m]=1;
return;
}*/
for(int i=2;i*i<=n;++i){
if(n%i==0){
p[++m]=i;c[m]=0;
while(n%i==0)n/=i,c[m]++;
}
}
if(n>1)p[++m]=n,c[m]=1;
return;
}
inline LL ksm(int a,int b){
LL cnt=1;
while(b){
if(b&1)cnt=(LL)cnt*a%mod;
a=(LL)a*a%mod;
b>>=1;
}
return cnt;
}
inline int sum(LL p,LL c){
if(c==0)return 1;
if(c%2==1)return((1+ksm(p,(c+1)/2))*sum(p,(c-1)/2))%mod;
if(c%2==0)return((1+ksm(p,c/2))*sum(p,(c/2)-1)+ksm(p,c))%mod;
}
int main(){
while(scanf("%d%d",&A,&B)!=EOF){
//if(A==0||A==1){printf("%d
",A);continue;}
//if(B==0){puts("1");continue;}
int ans=1;Divide(A);
for(int i=1;i<=m;++i)
ans=(ans*sum(p[i],B*c[i]))%mod;
printf("%d
",ans);
}
return 0;
}