BestCoder Round #6（1003）hdu4983（欧拉函数）

Goffi and GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 412    Accepted Submission(s): 34

Problem Description

    
    
    
    

     
     
     
     Goffi is doing his math homework and he finds an equality on his text book: 
     
     
     
     
     
     
     
      gcd(n−a,n)×gcd(n−b,n)=nk .
     
     
     
     

     
     
     
     
Goffi wants to know the number of (
     
     
     
     
     
     
     
      a,b ) satisfy the equality, if 
     
     
     
     
     
     
     
      n  and 
     
     
     
     
     
     
     
      k  are given and 
     
     
     
     
     
     
     
      1≤a,b≤n .
     
     
     
     

     
     
     
     
Note: 
     
     
     
     
     
     
     
      gcd(a,b)  means greatest common divisor of 
     
     
     
     
     
     
     
      a  and 
     
     
     
     
     
     
     
      b .
    
    
    
    

 

Input

    
    
    
    

     
     
     
     Input contains multiple test cases (less than 100). For each test case, there's one line containing two integers 
     
     
     
     
     
     
     
      n  and 
     
     
     
     
     
     
     
      k  (
     
     
     
     
     
     
     
      1≤n,k≤109 ).
    
    
    
    

 

Output

    
    
    
    

     
     
     
     For each test case, output a single integer indicating the number of (
     
     
     
     
     
     
     
      a,b ) modulo 
     
     
     
     
     
     
     
      109+7 .
    
    
    
    

 

Sample Input

    
    
    
    

     
     
     
     2 1
3 2
    
    
    
    

 

Sample Output

    
    
    
    

     
     
     
     2
1

     
     
     
     
 
      
 
       Hint 
      
 For the first case, (2, 1) and (1, 2) satisfy the equality. 
     
    
    
    
    

题意：RT

思路：很容易看出k>2时无解，k=2时一个解，k=1时可以预处理出G（x），表示gcd(i,n)=x的i的个数，这里求G（x）等价于求小于n/x且与之互素的数的个数，然后最后累加

            G（x）*G（n/x）即可，其中x是n的因子