POJ 1845 Sumdiv

Sumdiv

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).

 

 

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

两个重要的结论：

约数和公式：

对于已经分解的整数A=(p1^k1)*(p2^k2)*(p3^k3)*....*(pn^kn)

有A的所有因子之和为

    S = (1+p1+p1^2+p1^3+...p1^k1) * (1+p2+p2^2+p2^3+….p2^k2) * (1+p3+ p3^3+…+ p3^k3) * .... * (1+pn+pn^2+pn^3+...pn^kn)

 

用递归二分求等比数列1+pi+pi^2+pi^3+...+pi^n：

一、若n为奇数,一共有偶数项，则：
      1 + p + p^2 + p^3 +...+ p^n

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

     这里将其分成两半，提后一半的公因式p^(n/2+1)
      = (1 + p + p^2 +...+ p^(n/2)) * (1 + p^(n/2+1))

二、若n为偶数,一共有奇数项,则:
      1 + p + p^2 + p^3 +...+ p^n

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

     可知n-1为偶数，对前偶数项进行上述操作再加上最后一项p^n
      = (1 + p + p^2 +...+ p^(n/2-1)) * (1+p^(n/2)) + p^n;

 

AC代码如下：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define mod 9901#define M 10000#define ll long longusing namespace std;ll power(ll d,ll p)//幂次的优化{    ll ans=1;    while(p>0)    {        if(p%2)            ans=(ans*d)%mod;        p/=2;        d=(d*d)%mod;    }    return ans;}ll sum(ll d,ll p)//等比数列求和递归{    if(p==0)        return 1;    if(p==1)        return 1+d;    if(p%2==1)        return sum(d,p/2)*(1+power(d,p/2+1))%mod;    else return (sum(d,p/2-1)*(1+power(d,p/2))%mod+power(d,p))%mod;}int main(){    int i,j;    int a,b;    while(~scanf("%d%d",&a,&b))    {        int ds[M];        int po[M];        int tt=0;        for(i=2;i*i<=a;)//求a的因子        {            if(a%i==0)            {                ds[tt]=i;                po[tt]=0;                while(!(a%i))                {                    po[tt]++;                    a/=i;                }                tt++;            }            i==2?i++:i+=2;        }        if(a!=1)        {            ds[tt]=a;            po[tt++]=1;        }        ll ans = 1;        for(i=0;i<tt;i++)            ans=(ans*(ll)sum(ds[i],po[i]*b))%mod;        printf("%I64d\n",ans);    }}