求a^b的约数对mod取模

Posted 2021-01-25 033000-

const int maxn=30000+5;

int prime[maxn];
void marktable(int n){
    memset(prime,0,sizeof(prime));
    for(int i=2;i<=n;i++){
        if(!prime[i]) prime[++prime[0]]=i;
        for(int j=1;j<=prime[0]&&prime[j]<=n/i;j++){
            prime[prime[j]*i]=1;
            if(i%prime[j]==0) break;
        }
    }
}
long long factor[100][2];
int fatCnt;
int getFactors(long long x){
    fatCnt=0;
    long long tmp=x;
    for(int i=1;prime[i]<=tmp/prime[i];i++){
        factor[fatCnt][1]=0;
        if(tmp%prime[i]==0){
            factor[fatCnt][0]=prime[i];
            while(tmp%prime[i]==0){
                factor[fatCnt][1]++;
                tmp/=prime[i];
            }
            fatCnt++;
        }
    }
    if(tmp!=1) {
        factor[fatCnt][0]=tmp;
        factor[fatCnt++][1]=1;
    }
    return fatCnt;
}
int mod,A,B;
template<class T,class T1> T fast_mod(T a,T b,T1 Mod){
    a%=mod;
    if(b==0) return 1;
    T ans=1,base=a;
    while(b!=0){
        if(b&1)ans=(ans*base)%Mod;
        base=(base*base)%Mod;
        b>>=1;
    }
    return ans;
}

long long sum(long long p,long long n){
    if(p==0) return 0;
    if(n==0) return 1;
    if(n&1) return ((1+fast_mod(p,n/2+1,mod))%mod*sum(p,n/2)%mod)%mod;
    else return ((1+fast_mod(p,n/2+1,mod))%mod*sum(p,n/2-1)+fast_mod(p,n/2,mod)%mod)%mod;
}
long long solve(long long A,long long B){
    getFactors(A);
    long long ans=1;
    for(int i=0;i<fatCnt;i++){
        ans*=sum(factor[i][0],B*factor[i][1])%mod;
        ans%=mod;
    }
    return ans;
}