poj 1845 Sumdiv

Sumdiv

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 11880		Accepted: 2852

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).

参考思路：http://blog.csdn.net/lyy289065406/article/details/6648539

#include<stdio.h>
#include<math.h>
#include<string.h>
#define MOD 9901
long long p[100000],ans[100000];
long long Pow1(long long p,long long n)//计算p的n次方的函数，不能直接用pow函数
{
  /*   long long temp=p;//开始直接用循环，超时;
     int i;
     for(i=0;i<n-1;i++)
    temp=(temp%MOD*p%MOD)%MOD;

          //printf("temp=%lld
",temp);
 return temp;*/
long long  ret=1,s=p;//快速幂算法
    while(n)
    {
        if(n&1)
            ret=(ret%MOD*s%MOD)%MOD;
        if(n>>=1)
            s=(s%MOD*s%MOD)%MOD;
        else
            break;
    }
    return ret;
}
long long  Sum(long long  p,long long   n)//二分求等比数列的和
{
     if(n==0)
     return 1;
     else if(n%2!=0)
     return (((1+Pow1(p,n/2+1)%MOD))*(Sum(p,n/2))%MOD)%MOD;//两个公式，可以自己推下
     else
     return (((1+Pow1(p,n/2+1)%MOD)*(Sum(p,(n/2-1)))%MOD)+(Pow1(p,n/2))%MOD)%MOD;
}
int main()
{
     long long i,j,k,cnt,sum;
     int A,B;
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
    while(scanf("%d %d",&A,&B)!=EOF)
     {
          memset(ans,0,sizeof(ans));
            memset(p,0,sizeof(p));
            sum=1;
          k=0;
          for(i=2;i<=sqrt(A);i++)//把A分解
          {
               cnt=0;
               while(A%i==0)
               {

                    p[k]=i;
                    ans[k]++;
                    cnt=1;
                    A/=i;
               }
               if(cnt)
               k++;
          }
          /*
           if(A!=1)//这里很重要，我开始只考虑这个数是单个质数的情况，也就是A=13这种情况，
           {
           p[0]=A;//WA了几次，还有一种情况就是A=13*17这种如果我这样做的话，13的话就被覆盖了
           ans[0]=1;
           k=1;
           }*/
          if(A!=1)
          {
               p[k]=A;
               ans[k]+=1;
               k++;
          }
          for(j=0;j<k;j++)
          {

                sum=(sum%MOD*Sum(p[j],ans[j]*B)%MOD)%MOD;

          }

          printf("%lld
",sum%MOD);
     }

     return 0;
}