The number of divisors(约数) about Humble Numbers（约数个数公式）

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本文介绍了一种计算质朴数（仅由2、3、5、7质因数构成的数）的因数数量的方法，并提供了一个C++程序示例。通过分解质因数并利用因数个数公式，可以高效地得出答案。

The number of divisors(约数) about Humble Numbers

 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

裸的因数个数公式

code：

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