poj1845 Sumdiv 数论

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求A^B的所有约数之和
#include<cstdio>#include<cmath>#include<iostream>#include<algorithm>#include<cstring>using namespace std;#define MOD 9901const int MAXN=10000;int prime[MAXN+1];void getPrime(){   memset(prime,0,sizeof(prime));    for(int i=2;i<=MAXN;i++)    {        if(!prime[i])prime[++prime[0]]=i;        for(int j=1;j<=prime[0]&&prime[j]<=MAXN/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;}long long pow_m(long long a,long long n)//快速模幂运算{    long long res=1;    long long tmp=a%MOD;    while(n)    {        if(n&1){res*=tmp;res%=MOD;}        n>>=1;        tmp*=tmp;        tmp%=MOD;    }    return res;}long long sum(long long p,long long n)//计算1+p+p^2+````+p^n{    if(p==0)return 0;    if(n==0)return 1;    if(n&1)//奇数    {        return ((1+pow_m(p,n/2+1))%MOD*sum(p,n/2)%MOD)%MOD;    }    else return ((1+pow_m(p,n/2+1))%MOD*sum(p,n/2-1)+pow_m(p,n/2)%MOD)%MOD;}int main(){    int A,B;    getPrime();    while(scanf("%d%d",&A,&B)!=EOF)    {        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;        }        printf("%I64d\n",ans);    }    return 0;}
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