hdu 4983 Goffi and GCD（欧拉函数）

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

 

Sample Output

Hint

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

 

Source

BestCoder Round #6

 发现自己欧拉函数都给忘记了，所有赶紧补题。。。

1、k!=1时情况很简单，记住将if(k==2 || n==1)这个特判放在if(k>2)的前面，因为这个WA了很久，各种原因自己思考。

2、下面讨论k=1时情况。x=gcd(n-a,n),则n/x=gcd(n-b,n),因为n-a可以取到0...n-1也就是1....n，所以完全可以去掉n-这个限制条件，即gcd(a,n)=x、gcd(b,n)=n/x时个数，因为a<=n，所以gcd(a,n)的个数=u[n/x],u是欧拉函数。所以原式等于sigma(u[n/x]*u[x])其中x是n的约数。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 using namespace std;
 #define MOD 1000000007
 #define ll long long
 ll eular(ll n)
 {
     ll res=;
     for(ll i=;i*i<=n;i++)
     {
         if(n%i==)
         {
             n/=i,res*=i-;
             while(n%i==)
             {
                 n/=i;
                 res*=i;
             }
         }
     }
     if(n>) res*=n-;
     return res;
 }
 ll n,k;
 int main()
 {
     while(scanf("%I64d%I64d",&n,&k)==)
     {
         if(k== || n==)
         {
             printf("1\n");
             continue;
         }
         if(k>)
         {
             printf("0\n");
             continue;
         }

         ll ans=;
         for(ll i=;i*i<=n;i++)
         {
             if(n%i==)
             {
                 if(i*i!=n)
                   ans=(ans+eular(n/i)*eular(i)*)%MOD;
                 else
                     ans=(ans+eular(n/i)*eular(i))%MOD;
             }
         }
         printf("%I64d\n",ans);

     }
     return ;
 }