hdu 1492The number of divisors(约数) about Humble Numbers（因子个数）

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): 3344    Accepted Submission(s): 1638

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

题意：给你一个数n，求出n有多少个因子。比如4:1,2,4               12：1,2,3,4,6,12

思路：任何数都可以拆解成质数的多次方的乘积。且根据乘法原理，因子数=（1+e1）*（1+e2）...*（1+ek）   其中ei是第i个质因数的个数

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
#define LL long long
LL Devie(LL n)
{
    LL ans=1;
    for(LL i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            int num=1;
            while(n%i==0)
            {
                n/=i;
                num++;
            }
            ans*=num;
        }
    }
    if(n>1) return ans*2;
    return ans;
}
int main()
{
    LL n;
    while(~scanf("%lld",&n)&&n)
    {
        LL ans=Devie(n);
        printf("%lld\n",ans);
    }
    return 0;
}