Divisor number

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.

InputThe 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.

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

Sample Input
4
12
0
Sample Output
3
6
题意：这道题就是求一个数的因子个数，只是这个数的范围很大；
解题思路：
1.最开始的思路：这无非就是一个循环解决的事，应为想着循环的范围就在2到根号n，以为这样就行了，而且代码也简单，但是事实告诉我，其实在很大数的n的开根号之后还是很大，所以就超时了；

2.AC解题思路：这里用到了一个公式，是整数的唯一分解，比如，n可以分解为n=(p1^k1) * (p2^k2)… (pn^kn)；然后因子个数就为(1+k1)*(1+k2)…(1+kn)，这样就解决了超时问题；

#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
    long long n,i;
    while(~scanf("%lld",&n)&&n)
        {
            long long sum=1;
            for(i=2;i<=n;i++)
                {
                    if(n%i==0)
                        {
                            long long cnt=0;
                            while(n%i==0)
                                {
                                    cnt++;
                                    n=n/i;
                                }
                            sum=sum*(1+cnt);
                        }
                }
            printf("%lld\n",sum);
        }
    return 0;
}