ACM--约数--HDOJ 1492--The number of divisors(约数) about Humble Numbers

HDOJ题目地址：传送门

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): 3331    Accepted Submission(s): 1630

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
   
   
   
   
   
   
   
   


   
   
   
   

题意：给出一个只包含质因数2，3，5，7的数，求其约数的个数

思路：如果n=p1^n1*p2^n2*...*pn^nn,其中p1,p2,...,pn表示质因数，n1,n2,...,nn表示相应质因数的指数，根据乘法原理，则约数的个数为(n1+1)(n2+1)...(nn+1)

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main(){
   __int64 n,a,b,c,d;
   while(scanf("%I64d",&n)&&n){
      a=1;b=1;c=1;d=1;
      while(n!=1&&n%2==0){
         a++;
         n/=2;
      }
      while(n!=1&&n%3==0){
         b++;
         n/=3;
      }
      while(n!=1&&n%5==0){
         c++;
         n/=5;
      }
      while(n!=1&&n%7==0){
         d++;
         n/=7;
      }
      printf("%I64d\n",a*b*c*d);
   }

}