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The number of divisors(约数) about Humble Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3501 Accepted Submission(s): 1730

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

Sample Output

题解：假设要求的数是n:n = 2^a+3^b+5^c+7^d;要求其约数的个数，2，3，5，7每个因子都有（0到a,b,c,d)个选择，所以一共有（1+a)(1+b)(1+c)(1+d)个因子

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int div(long long int x,int m)
{
int ans = 0;
while(x%m==0)
{
  x/=m;
  ans++;
}
return ans;
}
int main()
{
    long long int n;
    while(cin >> n&&n)
    {
   long long int mu = 1;
   mu *= (div(n,2)+1);
   mu *= (div(n,3)+1);
   mu *= (div(n,5)+1);
   mu *= (div(n,7)+1);
   cout << mu << endl;
}
    return 0;
}

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