poj1845 数论 快速幂

Sumdiv

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 16466		Accepted: 4101

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). 

要求的是A^B的所有因子的和之后再mod 9901的值。

(1+a1+a1^2+...a1^n1)*(1+a2+a2^2+...a2^n2)*(1+a3+a3^2+...a3^n2)*...(1+am+am^2+...am^nm) mod 9901。

对于每一个(1+a1+a1^2+...a1^n1) mod 9901

等于 (a1^(n1+1)-1)/(a1-1) mod 9901，这里用到逆元的知识：a/b mod c = (a mod (b*c))/ b

所以就等于(a1^(n1+1)-1)mod (9901*(a1-1)) / (a1-1)。

至于前面的a1^(n1+1)，快速幂。

 #include<cstdio>
 #include<cmath>
 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #define mod 9901
 #define N 10007
 #define ll long long
 using namespace std;

 int prime[];

 void getPrime()
 {
     for(int i=;i<=;i++)
     {
         if(!prime[i])prime[++prime[]]=i;
         for(int j=;j<=prime[]&&prime[j]*i<=;j++)
         {
             prime[prime[j]*i]=;
             if(i%prime[j]==) break;
         }
     }
 }

 long long factor[][];
 int fatCnt;
 int getFactors(long long x)
 {
     fatCnt=;
     long long tmp=x;
     for(int i=;prime[i]<=tmp/prime[i];i++)
     {
         factor[fatCnt][]=;
         if(tmp%prime[i]==)
         {
             factor[fatCnt][]=prime[i];
             while(tmp%prime[i]==)
             {
                 factor[fatCnt][]++;
                 tmp/=prime[i];
             }
             fatCnt++;
         }
     }
     if(tmp!=)
     {
         factor[fatCnt][]=tmp;
         factor[fatCnt++][]=;
     }
     return fatCnt;
 }
 long long pow_m(long long a,long long n)//快速模幂运算
 {
     ll ans=;a%=mod;
     while(n)
     {
         if (n&) ans=(ans*a)%mod;
         n>>=;
         a=(a*a)%mod;
     }
     return ans;
 }
 long long sum(long long p,long long n)//计算1+p+p^2+````+p^n
 {
     if(p==)return ;
     if(n==)return ;
     if(n&) return ((+pow_m(p,n/+))%mod*sum(p,n/)%mod)%mod;
     else return ((+pow_m(p,n/+))%mod*sum(p,n/-)+pow_m(p,n/)%mod)%mod;

 }
 int main()
 {
     int A,B;
     getPrime();
     while(~scanf("%d%d",&A,&B))
     {
         getFactors(A);
         long long ans=;
         for(int i=;i<fatCnt;i++)
             ans=(ans*sum(factor[i][],B*factor[i][])%mod)%mod;
         printf("%lld\n",ans);
     }
 }