poj 1845 Sumdiv 约数和定理

Sumdiv

题目连接：

http://poj.org/problem?id=1845

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

题意

给你A,B,求A^B的因子和mod 9901

题解：

首先我们知道A的因式分解

A = (p1^k1) * (p2^k2) * (p3^k3) * .... * (pn^kn)

所以A^B = (p1^(k1*B)) * (p2^(k2*B)) * (p3^(k3*B)) * .... * (pn^(kn*B))

然后根据约数和定理，约数的和

Sum = (1+p1+p1^2+...+p1(k1*B))(1+p2....+p2^{(k2*B)).....(1+pn+...+pn}(kn*B))

中间等比数列要mod，所以就直接递归求就好了。

代码

#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<iostream>
using namespace std;
const int maxn = 1e6;
long long quickpow(long long  m,long long n,long long k)
{
    long long b = 1;
    while (n > 0)
    {
          if (n & 1)
             b = (b*m)%k;
          n = n >> 1 ;
          m = (m*m)%k;
    }
    return b;
}
int cnt[maxn];
int num[maxn];
int tot = 0;
void factorization(int x)
{
    for(int i=2;i*i<=x;i++)
    {
        if(x%i==0)
        {
            cnt[tot]=i;
            num[tot]=0;
            while(x%i==0)
            {
                x/=i;
                num[tot]++;
            }
            tot++;
        }
    }
    if(x!=1)
    {
        cnt[tot]=x;
        num[tot]=1;
        tot++;
    }
}
long long Sum_of_geometric_progression(long long p,long long n,long long mod)
{
    if(n==0)return 1;
    if(n&1)
        return ((1+quickpow(p,n/2+1,mod))%mod*Sum_of_geometric_progression(p,n/2,mod)%mod)%mod;
    else
        return (quickpow(p,n/2,mod)+(1+quickpow(p,n/2+1,mod))%mod*Sum_of_geometric_progression(p,(n-1)/2,mod)%mod)%mod;
}
int main()
{
    int A,B;
    while(scanf("%d%d",&A,&B)!=EOF)
    {
        tot = 0;
        factorization(A);
        int ans = 1;
        for(int i=0;i<tot;i++)
            ans = (ans*Sum_of_geometric_progression(cnt[i],B*num[i],9901))%9901;
        printf("%d\n",ans);
    }
    return 0;
}