POJ 1845 Sumdiv

快速幂+等比数列求和。。。。

                                                                                                  Sumdiv

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 12599		Accepted: 3057

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

#include <iostream> #include <cstdio> #include <cstring> using  namespace std; const  int MOD= 9901; typedef  long  long  int LL; int p[ 10000],n[ 10000],k,A,B; LL power(LL p,LL n)   ///p^n {     LL ans= 1;      while(n> 0)     {          if(n& 1)             ans=(ans*p)%MOD;         n>>= 1;         p=(p*p)%MOD;     }      return ans%MOD; } LL Spower(LL p,LL n)  ///1+p^1+p^2+p^3+....+p^n-1+p^n {      if(n== 0)  return  1;      if(n& 1)          return ((Spower(p,n/ 2)%MOD)*(( 1+power(p,n/ 2+ 1))%MOD))%MOD;      else          return (((Spower(p,n/ 2- 1)%MOD)*( 1+power(p,n/ 2+ 1))%MOD)%MOD+power(p,n/ 2)%MOD)%MOD; } int main() {      while(scanf( "%d%d",&A,&B)!=EOF)     {         k= 0;          for( int i= 2;i*i<=A;)         {              if(A%i== 0) p[k]=i,n[k]= 0,k++;              while(A%i== 0)             {                 n[k- 1]++;                 A/=i;             }              if(i== 2) i++;              else i+= 2;         }          if(A!= 1)         {             p[k]=A;             n[k++]= 1;         }         LL ans= 1;          for( int i= 0;i<k;i++)         {             ans=(ans*Spower(p ,B*n))%MOD;         }         printf("%I64d\n",ans);     }     return 0; }

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