Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1708 Accepted Submission(s): 833
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
1 #include<stdio.h>
2 int main()
3 {
4 long long n;
5 int p2,p3,p5,p7;
6 while(scanf("%I64d",&n),n)
7 {
8 p2=p3=p5=p7=0;
9 while(n%2==0)
10 {
11 n/=2;
12 p2++;
13 }
14 while(n%3==0)
15 {
16 n/=3;
17 p3++;
18 }
19 while(n%5==0)
20 {
21 n/=5;
22 p5++;
23 }
24 while(n%7==0)
25 {
26 n/=7;
27 p7++;
28 }
29 printf("%d\n",(p2+1)*(p3+1)*(p5+1)*(p7+1));//乘法原理，2 3 5 7可以一个都不要所以必须加一
30 }
31 return 0;
32 }
作者：火星十一郎
