Los Valley - P1593 - factor - Fermat's Little Theorem

Similar modulus as well as a relatively small pit that Lucas Theorem, and also the inverse would sometimes does not exist, except as a whole. By using some other method to avoid the inverse.

https://www.luogu.org/fe/problem/P1593
pit. You must be a good understanding of the conditions Fermat's little theorem inverse exist. Fermat's little theorem inversing the proviso that p is a prime number, and a is not 0, the Extended Euclidean Algorithm proviso that a, m coprime.
Then the above inversion yuan denominator geometric series with Fermat's Little Theorem, when there is no judge is not a zero. And they are not relatively prime so it can not use the extended Euclidean algorithm.
In fact, when a is reduced to 0 when the arithmetic sequence.

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int mod=9901;

ll a,b;
ll factor[100][2];
int ftop=0;

void fj() {
    for(ll i=2; i*i<=a; i++) {
        if(a%i==0) {
            ftop++;
            factor[ftop][0]=i;
            factor[ftop][1]=0;
            while(a%i==0) {
                factor[ftop][1]++;
                a/=i;
            }
        }
    }
    if(a!=1) {
        ftop++;
        factor[ftop][0]=a;
        factor[ftop][1]=1;
    }
    for(int i=1; i<=ftop; i++) {
        factor[i][1]*=b;
    }
    /*for(int i=1; i<=ftop; i++) {
        printf("%lld %lld\n",factor[i][0],factor[i][1]);
    }*/
}

ll da;

ll qpow(ll x,ll n) {
    x%=mod;
    ll res=1;
    while(n) {
        if(n&1) {
            res*=x;
            if(res>=mod)
                res%=mod;
        }
        x*=x;
        if(x>=mod)
            x%=mod;
        n>>=1;
    }
    if(res>=mod)
        res%=mod;
    return res;
}

void calc() {
    ll pro=1;
    for(int i=1; i<=ftop; i++) {
        ll sum=0;
        ll &p=factor[i][0];
        ll &k=factor[i][1];

        if((p-1)%mod==0) {
            //退化为等差数列
            sum=k+1;
        } else {
            sum=(qpow(p,k+1)-1+mod)%mod;
            //printf("sum=%lld\n",sum);
            sum*=qpow(p-1,mod-2);
        }
        if(sum>=mod)
            sum%=mod;
        //printf("sum=%lld\n",sum);

        pro*=sum;
        if(pro>=mod)
            pro%=mod;
    }

    da=pro%mod;
}

int main() {
#ifdef Yinku
    freopen("Yinku.in","r",stdin);
#endif // Yinku
    scanf("%lld%lld",&a,&b);
    if(a==0) {
        printf("0\n");
        return 0;
    }
    fj();
    calc();
    printf("%lld\n",da);
}