The number of divisors(约数) about Humble Numbers HDU - 1492 （唯一分解定理）

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本文介绍了一种计算特定类型数——谦逊数的因数数量的方法。谦逊数仅包含2、3、5或7作为其质因数。通过使用唯一分解定理，我们提供了一个高效的算法，用于确定任意给定谦逊数的因数总数。

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：

给一个数字n，求出n有多少个因数。

思路：

唯一分解定理：

一个大于1的正整数N，如果它的标准分解式为：，那么它的正因数个数为
N（n)=(1+a1)(1+a2)(1+a3)…(1+an)。

代码：

#include<stdio.h>
int main()
{
    long long n;
    while(~scanf("%lld",&n)&&n)
    {
        int a=1,b=1,c=1,d=1;
        while(n%2==0)
        {
            a++;
            n=n/2;
        }
        while(n%3==0)
        {
            b++;
            n=n/3;
        }
        while(n%5==0)
        {
            c++;
            n=n/5;
        }
        while(n%7==0)
        {
            d++;
            n=n/7;
        }
        printf("%d\n",a*b*c*d);
    }
    return 0;
}