POJ 2154 Color

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

这题不用考虑翻转操作

所以答案就是$ans=\sum_{i=1}^nn^{gcd(i,n)}$

枚举gcd，可以得到：

$ans=\frac{1}{n}\sum_{d=1}^n\sum_{i=1}^{n}n^{d}[gcd(i,n)==d]$
$ans=\sum_{d=1}^nn^{d-1}\sum_{i=1}^{\frac{n}{d}}[gcd(i,n/d)==1]$
$ans=\sum_{d=1}^nn^{d-1}\phi(\frac{n}{d})$

外层枚举n因数

求欧拉函数用定义式可以O(√n)求解

这样常数很大

实现筛出素数，求欧拉时只枚举素数

因为n的因数远远小于√n，所以很快，虽然理论上会超

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long lol;
int tot,vis[50001],prime[50001],ans,Mod;
void pre()
{int i,j;
  for (i=2;i<=50000;i++)
    {
      if (vis[i]==0)
    {
      tot++;
      prime[tot]=i;
    }
      for (j=1;j<=tot;j++)
    {
      if (1ll*i*prime[j]>50000) break;
      vis[i*prime[j]]=1;
      if (i%prime[j]==0)
        break;
    }
    }
}
int qpow(int x,int y)
{
  int res=1;
  while (y)
    {
      if (y&1) res=1ll*res*x%Mod;
      x=1ll*x*x%Mod;
      y>>=1;
    }
  return res;
}
int phi(int n)
{int i;
  int s=n;
  for (i=1;i<=tot&&1ll*prime[i]*prime[i]<=n;i++)
    if (n%prime[i]==0)
      {
    s=s-s/prime[i];
    while (n%prime[i]==0) n/=prime[i];
      }
  if (n!=1)
    s=s-s/n;
  return s%Mod;
}
int main()
{int T,tim,i,n;
  cin>>T;
  pre();
  while (T--)
    {
      scanf("%d%d",&n,&Mod);
      tim=sqrt(n);
      ans=0;
      for (i=1;i<=tim;i++)
    if (n%i==0)
      {
      ans=(ans+1ll*qpow(n,i-1)*phi(n/i)%Mod)%Mod;
      if (i*i<n)
        ans=(ans+1ll*qpow(n,n/i-1)*phi(i)%Mod)%Mod;
    }
      printf("%d\n",ans);
    }
}