description

solution

分解質因數然後等比數列求和，可能沒有逆元（本程式沒有處理這種情況），這時候需要特殊處理。

code

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define LF double
#define LL long long
#define ULL unsigned int
#define fo(i,j,k) for(LL i=j;i<=k;i++)
#define fd(i,j,k) for(LL i=j;i>=k;i--)
 

#define fr(i,j) for(LL i=begin[j];i;i=next[i])
using namespace std;
LL const mn=1e6+9,mo=9901;
LL n,m,a[mn],b[mn];
LL Pow(LL x,LL y){
    LL z=1;x%=mo;
    while(y){
        if(y&1)z=z*x%mo;
        x=x*x%mo;
        y>>=1;
    }
    return z;
}
int main(){
    //freopen("spring.in","r",stdin);
    //freopen
 
("spring.out","w",stdout);
    freopen("d.in","r",stdin);
    freopen("d.out","w",stdout);
    scanf("%lld%lld",&n,&m);
    LL lim=sqrt(n);
    fo(i,2,lim)if(n%i==0){
        a[++a[0]]=i;
        while(n%i==0)b[a[0]]++,n/=i;
    }
    if(n!=1)a[++a[0]]=n,b[a[0]]=1;
    LL ans=1;
    fo(i,1,a[0])ans=ans*(
 
Pow(a[i],b[i]*m+1)-1)%mo*Pow(a[i]-1,mo-2)%mo;
    printf("%lld",(ans+mo)%mo);
    return 0;
}