【POJ 2154】Color（置换群）

Color
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 9265 Accepted: 3010

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

[题意][将正n边形的n个顶点用n种颜色染色，问有多少种方案（答案mod p，且可由旋转互相得到的算一种）]

【题解】【与前面几题相似，但此题没有旋转，但数据范围相当恶心，所以，要优化】
∑ni=1ngcd(i,n)
=∑ni=1∑nd=1n[(i,n)=d]
=∑[d|n]∗φ(d n)∗nd
按照这个公式筛欧拉函数，但因为范围过大，所以，不可能全部筛出，筛一半，另一半，用下面的方法

根号n的时间内求一个数的phi
inline int find(int m)
{
    int i,k=m,l=m;
    for (i=2;i*i<=m;++i)
     if (!(l%i))
      {
        k=k-k/i;
        do{  l/=i;  }while(!(l%i));
      }
    if (l>1) k=k-k/l;
    return k;
}

由于范围过大，枚举的时候只能枚举一半，而且要谨慎使用long long

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
int prime[100010],phi[100010];
int m,n,mod;
bool p[100010];

inline void shai(int m)
{
    int i,j;
    phi[1]=1; p[1]=1;
    for(i=2;i<=m;++i)
      {
        if(!p[i])  prime[++prime[0]]=i,phi[i]=i-1;
        for(j=1;j<=prime[0];++j)
         {
            if(i*prime[j]>m) break;
            p[i*prime[j]]=1;
            if(!(i%prime[j])) 
              {phi[i*prime[j]]=phi[i]*prime[j]; break;}
             else phi[i*prime[j]]=phi[i]*(prime[j]-1);
          } 
      }
    return;
}
ll poww(int x,int q)
{
    if(!q) return 1;
    if(q==1) return x%mod;
    if(q==2) return x*x%mod;
    if(q%2)
      {ll sum=poww(x,q/2)%mod;  sum=sum*sum*x%mod;  return sum%mod;}
     else
      {ll sum=poww(x,q/2)%mod; sum*=sum; return sum%mod; }
}
inline ll get_phi(int x)
{
    if(x<=70000) return phi[x];
    ll ans=x;
    int i;
    for(i=2;i*i<=x;++i)
     if(!(x%i))
      {
        ans=ans*(i-1)/i;
        while(!(x%i)) x/=i;
      }
    if(x>1) ans=ans*(x-1)/x;
    return ans%mod;
}

int main()
{
    int i,j;
    shai(100000);
    scanf("%d",&n);
    for(j=1;j<=n;++j)
     {
        ll ans=0;
        scanf("%d%d",&m,&mod);
        for(i=1;i*i<=m;++i) 
          if(!(m%i))
           {
            ll s1=get_phi(i); ll s2=get_phi(m/i);
            if(m/i==i) ans=(ans+s2*poww(m%mod,i-1))%mod;
             else ans=(ans+s2*poww(m%mod,i-1))%mod,
            ans=(ans+s1*poww(m%mod,m/i-1))%mod; 
           }
        printf("%I64d\n",ans%mod);
     }
    return 0;
}