hdu 3923 Invoker c中颜色染n个格子，旋转和翻转算一种

 

Problem Description

On of Vance's favourite hero is Invoker, Kael. As many people knows Kael can control the elements and combine them to invoke a powerful skill. Vance like Kael very much so he changes the map to make Kael more powerful.

In his new map, Kael can control n kind of elements and he can put m elements equal-spacedly on a magic ring and combine them to invoke a new skill. But if a arrangement can change into another by rotate the magic ring or reverse the ring along the axis, they will invoke the same skill. Now give you n and m how many different skill can Kael invoke? As the number maybe too large, just output the answer mod 1000000007.

 

 

Input

The first line contains a single positive integer T( T <= 500 ), indicates the number of test cases.
For each test case: give you two positive integers n and m. ( 1 <= n, m <= 10000 )

 

 

Output

For each test case: output the case number as shown and then output the answer mod 1000000007 in a line. Look sample for more information.

 

 

Sample Input

  
  

   
   2
3 4
1 2
  
  

 

 

Sample Output

  
  

   
   Case #1: 21
Case #2: 1


   
   

    
    

     
     Hint
    
    
For Case #1: we assume a,b,c are the 3 kinds of elements.
Here are the 21 different arrangements to invoke the skills
/ aaaa / aaab / aaac / aabb / aabc / aacc / abab / 
/ abac / abbb / abbc / abcb / abcc / acac / acbc /
/ accc / bbbb / bbbc / bbcc / bcbc / bccc / cccc /
   
   
  
  

 

//

 

#include <cstdio>
const int mod=1000000007;
int fun(int a,int n)
{
    if(n==1) return a%mod;
    if(n==0) return 1;
    long long t=fun(a,n/2);
    t=(t*t)%mod;//important
    if(n&1) return (t*a)%mod;
    else return t%mod;
}
int euler(int n)
{
    int ans = n;
    for (int i = 2; i <= n; i++) if (n % i == 0)
        {
            ans -= ans / i;
            while (n % i == 0) n /= i;
        }
    if (n > 1) ans -= ans / n;
    return ans;
}
//求(a/b)%p,gcd(B,p)=1
__int64 exgcd(__int64 a,__int64 b,__int64& x,__int64& y)
{
    if(b==0) return x=1,y=0,a;
    __int64 r=exgcd(b,a%b,x,y);
    __int64 t=x;
    x=y;
    y=t-(a/b)*y;
    return r;
}
__int64 calc(__int64 a,__int64 b,__int64 mod)
{
    __int64 x,y;
    __int64 r=exgcd(b,mod,x,y);
    x*=a;
    x=(x%mod+mod)%mod;
    return x;
}
int main()
{
    int n, c;
    int cas = 1, T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &c, &n);
        if (n == 0)
        {
            printf("0\n");
            continue;
        }
        __int64 ans = 0;
        for (int i = 1; i <= n; i++)
        if (n % i == 0)
            {
                ans += (long long)fun(c, i) * euler(n / i);
                ans %= mod;
            }//枚举个数
        if (n & 1)
        {
            ans += (long long)n * fun(c, n / 2 + 1);
            ans %= mod;
        }
        else
        {
            ans += (long long)n / 2 * (fun(c, n / 2) + fun(c, n / 2 + 1));
            ans %= mod;
        }
//        ans /= (2 * n);
//        ans %= mod;
        ans=calc(ans,(long long)2*n,(long long)mod);
        printf("Case #%d: %I64d\n",cas++, ans);
    }
    return 0;
}