POJ 2154 Color

Description

Beads of N colors are connected together into a circular necklace of N beads (N<=1000000000). Your job is to calculate how many different kinds of the necklace can be produced. You should know that the necklace might not use up all the N colors, and the repetitions that are produced by rotation around the center of the circular necklace are all neglected.
You only need to output the answer module a given number P.

Input

The first line of the input is an integer X (X <= 3500) representing the number of test cases. The following X lines each contains two numbers N and P (1 <= N <= 1000000000, 1 <= P <= 30000), representing a test case.

Output

For each test case, output one line containing the answer.

Sample Input

5
1 30000
2 30000
3 30000
4 30000
5 30000

Sample Output

1
3
11
70
629

Source

POJ Monthly,Lou Tiancheng

思路

首先利用polya求出答案为：

ans=1n∑i=0n−1ngcd(n,i)$$ans=\frac{1}{n}\sum _{i=0}^{n-1}{n}^{gcd(n,i)}$$

那么化简：

ans=====1n∑i=1nngcd(n,i)∑i=1nngcd(n,i)−1∑i=1n∑d=1n[gcd(n,i)==d]nd−1∑d|n∑i=1nd[gcd(nd,i)==1]nd−1∑d|nφ(nd)nd−1(66)(67)(68)(69)(70)$$\begin{array}{}\text{(66)}& ans& =& \frac{1}{n}\sum _{i=1}^n{n}^{gcd(n,i)}\text{(67)}& & =& \sum _{i=1}^n{n}^{gcd(n,i)-1}\text{(68)}& & =& \sum _{i=1}^n\sum _{d=1}^n[gcd(n,i)==d]n^{d-1}\text{(69)}& & =& \sum _{d|n}\sum _{i=1}^{\frac{n}{d}}[gcd(\frac{n}{d},i)==1]n^{d-1}\text{(70)}& & =& \sum _{d|n}\phi (\frac{n}{d})n^{d-1}\end{array}$$

时间复杂度O(n0.75)$O(n^{0.75})$。

代码

#include <cstdio>

inline int phi(int n)
{
  int res=n;
  for(register int i=2; i*i<=n; ++i)
    {
      if(!(n%i))
        {
          res=(res/i)*(i-1);
          while(!(n%i))
            {
              n/=i;
            }
        }
    }
  if(n!=1)
    {
      res=(res/n)*(n-1);
    }
  return res;
}

inline int quickpow(int a,int b,int p)
{
  int res=1;
  while(b)
    {
      if(b&1)
        {
          res=1ll*res*a%p;
        }
      a=1ll*a*a%p;
      b>>=1;
    }
  return res;
}

int t,n,p;

int main()
{
  scanf("%d",&t);
  while(t--)
    {
      int ans=0;
      scanf("%d%d",&n,&p);
      for(register int i=1; i*i<=n; ++i)
        {
          if(!(n%i))
            {
              ans+=1ll*phi(n/i)*quickpow(n,i-1,p)%p;
              if(i*i!=n)
                {
                  ans+=1ll*phi(i)*quickpow(n,n/i-1,p)%p;
                }
              ans%=p;
            }
        }
      printf("%d\n",ans);
    }
  return 0;
}