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): 3416 Accepted Submission(s): 1676
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
“2006校园文化活动月”之“校庆杯”大学生程序设计竞赛暨杭州电子科技大学第四届大学生程序设计竞赛
/*
求一个数的约数：x=p1^x1 * p2^x2 * p3^x3.....
p1p2p3分别为素数，那么x的约数的个数就是ans=(x1+1)*(x2+1)*(x2+1)....
*/
#include <set>
#include <map>
#include <cmath>
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long LL;
using namespace std;
int a[6]={2,3,5,7};
int main ()
{
LL n;
while(scanf("%lld",&n),n)
{
LL ans=1;
int i=0;
int t=0;
while(n)
{
if(n%a[i]==0)
{
t++;
n/=a[i];
}
else
{
ans*=(t+1);
t=0;
i++;
if(n==1)
break;
}
}
printf("%lld
",ans);
}
return 0;
}