POJ 1845 Sumdiv

快速幂+等比数列求和。。。。

                                                                                                 Sumdiv

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

DescriptionConsider 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).InputThe only line contains the two natural numbers A and B, (0 <= A,B <= 50000000)separated by blanks.OutputThe only line of the output will contain S modulo 9901.Sample Input2 3Sample Output15Hint2^3 = 8. 
The natural divisors of 8 are: 1,2,4,8. Their sum is 15. 
15 modulo 9901 is 15 (that should be output). 
SourceRomania 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[i]))%MOD;         }         printf("%I64d\n",ans);     }     return 0; } [/i]

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