约数

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.

InputThe 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.

OutputFor each test case, output its divisor number, one line per case.

Sample Input
4
12
0Sample Output
3
6
方法一：

求质因子只有2，3，5，7的整数有多少个约数，没啥解法，结论：如果一个数可以表达成num=a^x* b^y * c^z…，则num的约数个数为（a+1）（b+1）（c+1）…；

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

方法二：

#include <stdio.h>
#include <algorithm>
using namespace std;
typedef long long ll;
const ll N=1e6+10;
ll prime[N]={2,3,5,7};
ll xia(ll a)
{
 ll i,x,ans=1;
 for(i=0;i<4&&prime[i]<a;i++)
 {
  x=0;
  while(a%prime[i]==0)
  {
   x++;
   a/=prime[i];
  }
  ans*=x+1;
 }
 if(a>1)
 ans*=1+1;
 return ans;
}
int main()
{
 ll n;
 while(~scanf("%lld",&n)&&n)
 {
  ll s=xia(n);
  printf("%lld\n",s);
 }
}