SPOJ Equation :求 1/n!=1/x+1/y 的解的个数

Description

You are given integer positive number N. Find the number of solutions in positive integer numbers of the following equation:

1/N! = 1/X + 1/Y

Input

Each line of input file contains one integer number N (1 ≤ N ≤ 10⁴). The last line contains 0 and shouldn't be processed. Number of lines in the input does not exceed 30.

Output

For every line in the input write the answer on a separate line.

Sample Input

Input:
1
2
0

Output:
1
3

//题目就是求1/n! = 1/x + 1/y 的解的个数，看样例知要考虑(x,y)对数的关系。

//设 m=n! ，由等式知x,y必定大于n!，所以再设 x=n!+k=m+k 带入 1/m=1/x+1/y 中化简得到y=m*m/k+m，因y为整数，所以要求k整除m*m，即k为m*m的因子，问题便转化为求n!*n!的因子个数， 设n!=p1^e1 * p2^e2 * p3^e3 *...*pk^ek，则 n!*n!= p1^(2*e1) * p2^(2*e2) *...*pk^(2*ek) 。 则因子个数sum=(2*e1+1)*(2*e2+1)*...*(2*ek+1)； 答案很大，需要高精度。用java处理方便！

 import java.util.Scanner;
 import java.io.*;
 import java.math.*;
 import java.math.BigInteger; 
 
 public class Main
 {
      static boolean x[]=new boolean[10000+10];
      static int prime[]=new int[10000+10],num[]=new int[10000+10],cnt=0;
      public static void Init()
      {
          int i,tmp;
          for(i=0;i<=10000;i++) x[i]=false;
          x[0]=x[1]=true;
          for(i=2;i<=10000;i++)
               if(x[i]==false)
               {
                    prime[++cnt]=i;
                    tmp=i*i;
                    while(tmp<=10000)
                    {
                         x[tmp]=true;
                         tmp+=i;
                    }
               }
      }
      public static void main(String[] args)
      {
          Scanner cin=new Scanner(System.in);
          Init();
          int n,i,tmp;
          while(cin.hasNext())
          {
               n=cin.nextInt();
               if(n==0) break;
               for(i=0;i<=10000;i++) num[i]=0;
               for(i=1;i<=cnt&&prime[i]<=n;i++) //分解出n!内所有因子的个数
               {
                    tmp=n;
                    while(true)
                    {
                         if(tmp==0||(prime[i]>n))
                              break;
                         num[i]+=tmp/prime[i];
                         tmp/=prime[i];
                    }
               }
               BigInteger ans=BigInteger.ONE;
               for(i=1;i<=cnt&&prime[i]<=n;i++)
                    if(num[i]>=1)
                         ans=ans.multiply(BigInteger.valueOf(2*num[i]+1));
               System.out.println(ans);
          }
      }
 }