定理：一个正整数 n 可以用素因子唯一表示为 p1^r1 * p2^r2 * ...
pk^rk (其中 pi 为素数) , 那么这个数的因子的个数就是，(r1+1)*(r2+1)*...*(rk+1).

理解：为什么是加1之后再相乘，因为一个数的的因子数至少为1和他自身，但因为r1，r2。。可以为0，所以因子的个数为（r1+1）。。。

拓展一下：

定理1： 一个正整数 n 可以用素因子唯一表示为 p1^r1 * p2^r2 * ... pk^rk (其中 pi 为素数) , 那么这个数的因子的个数就是，(r1+1)*(r2+1)~(rk+1).

定理2：如果一个数字 n = p1^r1 * p2^r2 * ... pk^rk ，那么 n*n = p1^r1 * p2^r2 * ... pk^rk   * p1^r1 * p2^r2 * ... pk^rk ，它的因子的个数就是 (2*r1+1)*(2*r2+1)~(2*rk+1)

Humble Numbers只要2,3,5,7，因此只要统计这四个因子的数目就可以了！

The number of divisors(约数) about Humble Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

代码：

#include<stdio.h>
int main()
{
    __int64 n, c2, c3, c5, c7;
    while(scanf("%I64d", &n), n){
        c2 = c3 = c5 =c7 = 1;
        while(n%7 == 0){
            c7++;
            n/=7;
        }
        while(n%5==0){
            c5++;
            n/=5;
        }
        while(n%3==0){
            c3 ++;
            n /= 3;
        }
        while(n%2 == 0){
            c2 ++;
            n /= 2;
        }
        __int64 ans =c2*c3*c5*c7;
        printf("%I64d\n", ans);
    }
}

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