POJ 1845 Sumdiv（数论，求A^B的所有约数和）

Sumdiv

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 10071		Accepted: 2357

Description

Consider two natural numbers A and B. Let S be the sum of all natural divisors of A^B. Determine S modulo 9901 (the rest of the division of S by 9901).

Input

The only line contains the two natural numbers A and B, (0 <= A,B <= 50000000)separated by blanks.

Output

The only line of the output will contain S modulo 9901.

Sample Input

2 3

Sample Output

15

Hint

2^3 = 8.
The natural divisors of 8 are: 1,2,4,8. Their sum is 15.
15 modulo 9901 is 15 (that should be output).

Source

Romania OI 2002

 

 

 

题意:求 A^B的所有约数之和对9901取模后的结果.
    根据唯一分解定理将A进行因式分解可得:A = p1^a1 * p2^a2 * p3^a3 * pn^an.
    A^B=p1^(a1*B)*p2^(a2*B)*...*pn^(an*B);
    A^B的所有约数之和sum=[1+p1+p1^2+...+p1^(a1*B)]*[1+p2+p2^2+...+p2^(a2*B)]*[1+pn+pn^2+...+pn^(an*B)].

 

      等比数列1+pi+pi^2+pi^3+...+pi^n可以由二分求得(即将需要求解的因式分成部分来求解)
       若n为奇数,一共有偶数项,设p为3,则(1+p)+(p^2+p^3)=(1+p)+p^2(1+p)=(1+p^2)*(1+p)

                  1+p+p^2+p^3+........+p^n=(1+p+p^2+....+p^(n/2))*(1+p^(n/2+1));
       若n为偶数,一共有奇数项,设p为4,则(1+p)+p^2+(p^3+p^4)=(1+p)+p^2+p^3(1+p)=(1+p^3)*(1+p)+P^2

                  1+p+p^2+p^3+........+p^n=(1+p+p^2+....+p^(n/2-1))*(1+p^(n/2+1));

 

/*
POJ 1845 Sumdiv
求A^B的所有约数之和%9901

*/
#include<stdio.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;

#define MOD 9901

//******************************************
//素数筛选和合数分解
const 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()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    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;
}