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

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)。

#include <stdio.h>
using namespace std;
 typedef long long ll;
 int main()
{
    ll nn;
   while(~scanf("%lld",&nn) && nn)
{
    int b[] = {2,3,5,7};
    int a[] = {1,1,1,1};
    for(int i = 0; i < 4; i++){
        if(nn % b[i] == 0)
		{
            while(nn % b[i] == 0)
			{
                a[i]++;
                nn /= b[i];
            }
        }
    }
    printf("%d\n",a[0]*a[1]*a[2]*a[3]);
}
return 0;
 }