HDUOJ---The number of divisors(约数) about Humble Numbers

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): 2039    Accepted Submission(s): 1002

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
 
Author
lcy
 
Source
Recommend
LL
分解质因数
 1 //http://acm.hdu.edu.cn/showproblem.php?pid=1492
 2 #include<iostream>
 3 #include<cstdio>
 4 using namespace std;
 5 int main()
 6 {
 7     _int64 i,n,c1,c2,c3,c4;
 8     
 9     while(scanf("%I64d",&n),n)
10     { 
11         for(c1=1,i=n;i%2==0&&i!=0;i/=2) 
12                    c1++;
13         for(c2=1,i=n;i%3==0&&i!=0;i/=3)
14                     c2++;
15         for(c3=1,i=n;i%5==0&&i!=0;i/=5) 
16                     c3++;
17          for(c4=1,i=n;i%7==0&&i!=0;i/=7) 
18                     c4++;
19         printf("%I64d
",c1*c2*c3*c4);
20     }
21 return 0;
22 }
View Code
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原文地址:https://www.cnblogs.com/gongxijun/p/3242749.html