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POJ 1845 Sumdiv(数论)


Sumdiv


Time Limit: 1000MS

 

Memory Limit: 30000K

Total Submissions: 15745

 

Accepted: 3894


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). 


Source




     最近在学习数论的一些知识点,感觉无处下手,只好按照相应的题目进行学习,做这个题需要用到以下知识点。



(1)   整数的唯一分解定理:

      任意正整数都有且只有一种方式写出其素因子的乘积表达式。

      A=(p1^k1)*(p2^k2)*(p3^k3)*....*(pn^kn)   其中pi均为素数

(2)   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)

(3)   同余模公式:

(a+b)%m=(a%m+b%m)%m

(a*b)%m=(a%m*b%m)%m

 

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

using namespace std;

const int h = 10001;
int n,m;
struct node
{
    int x;
    int y;
}q[h];

long long int power(long long int pn,long long int pm)   ///反复平方法来求A^B  省时间
{
    long long int sq = 1;
    while(pm>0)
    {
        if(pm%2)
        {
            sq = (sq*pn)%9901;
        }
        pm = pm / 2;
        pn = pn * pn % 9901;
    }
    return sq;
}

long long int updata(long long int pn,long long int pm)  ///递归二分求等比数列的和
{
    if(pm == 0)
    {
        return 1;
    }
    if(pm%2)
    {
        return (updata(pn,pm/2)*(1+power(pn,pm/2+1)))%9901;  /// 当pm为奇数时,有公式来求等比数列的和  (1 + p + p^2 +...+ p^(n/2)) * (1 + p^(n/2+1)) 
    }
    else
    {
        return (updata(pn,pm/2-1)*(1+power(pn,pm/2+1)) + power(pn,pm/2))%9901;  ///当pm为偶数时,有公式来求等比数列的和  (1 + p + p^2 +...+ p^(n/2-1)) * (1+p^(n/2+1)) + p^(n/2); 
    }
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        int k = 0;
        for(int i=2;i*i<=n;)   ///寻找质因子,一个很好的方法
        {
            if(n%i == 0)
            {
                q[k].x = i;
                q[k].y = 0;
                while(n%i == 0)
                {
                    q[k].y++;
                    n /= i;
                }
                k++;
            }
            if(i == 2)
            {
                i++;
            }
            else
            {
                i = i + 2;
            }
        }
        if(n!=1)
        {
            q[k].x = n;
            q[k].y = 1;
            k++;
        }
        int ans = 1;
        for(int i=0;i<k;i++)
        {       
            ans = (ans*(updata(q[i].x,q[i].y*m)%9901)%9901);
        }
        printf("%d\n",ans);
    }
    return 0;
}




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