HDU 1492 The number of divisors(约数) about Humble Numbers(数论题目要知道定理呀....)

The number of divisors(约数) about Humble Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3339    Accepted Submission(s): 1633

Problem Description

A number whose only prime factors are 2,3,5 or 7 is called a humble number. The sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, ... shows the first 20 humble numbers. 

Now given a humble number, please write a program to calculate the number of divisors about this humble number.For examle, 4 is a humble,and it have 3 divisors(1,2,4);12 have 6 divisors.

 

Input

The input consists of multiple test cases. Each test case consists of one humble number n,and n is in the range of 64-bits signed integer. Input is terminated by a value of zero for n.

 

Output

For each test case, output its divisor number, one line per case.

 

Sample Input

   
   
   
   

    
    
    
    4
12
0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    3
6
   
   
   
   

 

Author

lcy

 

Source

“2006校园文化活动月”之“校庆杯”大学生程序设计竞赛暨杭州电子科技大学第四届大学生程序设计竞赛

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1492

题意:求素因子(2,3,5,7)的个数,

这道题目就是一道数学题，题意大致为问某个数是否为“谦虚数”，就是满足所有的因子都是由2,3,5,7这四个质因子所组成的数据，然后问这个数有多少个因子；

解法：就是将数据拆分，统计每个素因子的个数，然后直接相乘即可先分解质因数.

数论里的结论:
如果一个数分解质因数的形式是：M = x^a * y^b * z^c * ...

则M的约数个数 = (a+1)(b+1)(c+1)...

AC代码:

/**先分解质因数
如果一个数分解质因数的形式是：
M = x^a * y^b * z^c * ...
则M的约数个数 = (a+1)(b+1)(c+1)...*/

#include <iostream>
using namespace std;
int main()
{
    __int64 n,a,b,c,d;
    while(cin>>n,n)
    {
        a=b=c=d=1;
        while(n%2==0)
        {
            a++;
            n/=2;
        }
        while(n%3==0)
        {
            b++;
            n/=3;
        }
        while(n%5==0)
        {
            c++;
            n/=5;
        }
        while(n%7==0)
        {
            d++;
            n/=7;
        }
        cout<<a*b*c*d<<endl;
    }
    return 0;
}