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). 

/**
題意：求A^B的所有約數之和 Mod 9901。

思路：大數模運算。兩個最基本公式：(A*B)%C = ((A%C)*(B%C))%C 和 (A+B)%C = ((A%C)+(B%C))%C 。

1: 對A進行素因子分解得
     A = p1^a1 * p2^a2 * p3^a3 *...* pn^an.
  故 A^B = p1^(a1*B) * p2^(a2*B) *...* pn^(an*B);

2：A^B的所有約數之和爲：
     sum = [1+p1+p1^2+...+p1^(a1*B)] * [1+p2+p2^2+...+p2^(a2*B)] *...* [1+pn+pn^2+...+pn^(an*B)].
  如 200 = 2^3 * 5^2 :  sum(200) = [1 + 2 + 4 + 8] * [1 + 5 + 25].

3: 求等比數列1+pi+pi^2+pi^3+...+pi^n可以由遞歸形式的二分求得：（模運算不能用等比數列和公式！）
   若n爲奇數,一共有偶數項，則：
      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))
      = (1 + p + p^2 +...+ p^(n/2)) * (1 + p^(n/2+1))
   如：1 + p + p^2 + p^3 + p^4 + p^5 = (1 + p + p^2) * (1 + p^3)
   若n爲偶數,一共有奇數項,則:
      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)
      = (1 + p + p^2 +...+ p^(n/2-1)) * (1+p^(n/2+1)) + p^(n/2);
   如：1 + p + p^2 + p^3 + p^4 = (1 + p) * (1 + p^3) + p^2
**/
#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <vector>
#include <bitset>
#include <cstdio>
#include <string>
#include <numeric>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
typedef long long  ll;
typedef unsigned long long ull;

int dx[4]={-1,1,0,0};
int dy[4]={0,0,-1,1};//up down left right
bool inmap(int x,int y,int n,int m){if(x<1||x>n||y<1||y>m)return false;return true;}
int hashmap(int x,int y,int m){return (x-1)*m+y;}

#define eps 1e-8
#define inf 0x7fffffff
#define debug puts("BUG")
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)
#define Max 50000001
#define maxn 8008
#define mod  9901

int p[maxn];//Primer
int cnt[maxn];
int N,K;

ll Pow(ll n,ll k)//自寫快速冪
{
    ll ans=1;
    while(k>0)//要記住
    {
        if(k&1)
            ans=(ans*n)%mod;
        k>>=1;
        n=n*n%mod;
    }
    return ans;
}

ll sum(ll n,ll k)//遞歸求和
{
    if(k==0)
        return 1;
    if(k&1)
        return ((1+Pow(n,k/2+1))*sum(n,k/2))%mod;
    else
        return ((1+Pow(n,k/2+1))*sum(n,k/2-1)+Pow(n,k/2))%mod;
}

int main()
{
    while(~scanf("%d%d",&N,&K))
    {
        int k=0;
        for(int i=2;i*i<=N;i++)
        {
            if(N%i==0)
            {
                p[k]=i;//顯然，i一定是素數
                while(N%i==0)
                {
                    N/=i;
                    cnt[k]++;
                }//算出了基本算數定理中每一個素數的冪，N是不斷變化的，使得i能夠一直是素數
            }
            k++;
        }
        if(N!=1)//N有可能是素數，那前面的循環中的if就會一次也沒有執行
        {
            p[k]=N;
            cnt[k++]=1;
        }
        ll ans=1;
        for(int i=0;i<k;i++)
            ans=ans*(sum(p[i],cnt[i]*K)%mod)%mod;
        printf("%lld\n",ans);
    }
    return 0;
}