The number of divisors(约数) about Humble Numbers

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The number of divisors(约数) about Humble Numbers

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

根据这个式子可以知道n是可以被拆分为一系列素数的乘积，那么将这些素数任意组合（当然每个素数的使用次数是有限制的）的得到的数字必然也是n的因子，比如18=2*3*3,9=3*3,6=2*3，那么6和9都是18的因子，总结一下对于每个是n的因子的素数p（它的指数是e，有e+1种选择，即不选p，选一个p，选两个p，……选n个p）剩下的就是一个组合的问题，对于n的约数的个数就是各个是因子的素数的选择的数目的乘积

#include<stdio.h>
int main()
{long long n;while(scanf("%lld",&n),n){int an[]={2,3,5,7};long long ans[4]={0};for(int i=0;i<4;i++){while(n%an[i]==0){ans[i]++;n/=an[i];}}printf("%lld\n",(ans[0]+1)*(ans[1]+1)*(ans[2]+1)*(ans[3]+1));}return 0;
}

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