hdu 1492(约数的个数)

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): 3337    Accepted Submission(s): 1632

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

 

如果n = p1^a1p2^a2p3^a3...pn^an 那么根据乘法原理,约数的个数为 (a1+1)*(a2+1)...(an+1)

#include <stdio.h>
#include <math.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdlib.h>
using namespace std;
typedef long long LL;

int main()
{
    LL n;
    while(scanf("%lld",&n)!=EOF,n)
    {
        int cnt = 0;
        int a=0,b=0,c=0,d=0;
        if(n%2==0)
        {
            while(n%2==0)
            {
                n/=2;
                a++;
            }
        }
        if(n%3==0)
        {
            while(n%3==0)
            {
                n/=3;
                b++;
            }
        }
        if(n%5==0)
        {
            while(n%5==0)
            {
                n/=5;
                c++;
            }
        }
        while(n%7==0)
        {
            while(n%7==0)
            {
                n/=7;
                d++;
            }
        }
        printf("%lld
",(LL)(1+a)*(1+b)*(1+c)*(1+d));
    }
}