poj2154-color-polyan次二面体+欧拉函数优化

N<=1e9,O(nlogn)的做法会超时。从枚举置换转变为枚举轮换长度，然后可以利用欧拉函数，把复杂度变为O(√n * logn)

 /*--------------------------------------------------------------------------------------*/

 #include <algorithm>
 #include <iostream>
 #include <cstring>
 #include <ctype.h>
 #include <cstdlib>
 #include <cstdio>
 #include <vector>
 #include <string>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <set>
 #include <map>

 //debug function for a N*M array
 #define debug_map(N,M,G) printf("\n");for(int i=0;i<(N);i++)\
 {for(int j=;j<(M);j++){\
 printf("%d",G[i][j]);}printf("\n");}
 //debug function for int,float,double,etc.
 #define debug_var(X) cout<<#X"="<<X<<endl;
 #define LL long long
 const int INF = 0x3f3f3f3f;
 const LL LLINF = 0x3f3f3f3f3f3f3f3f;
 /*--------------------------------------------------------------------------------------*/
 using namespace std;

 int N,M,T;
 int MOD = 1e9+;

 int Eular(int n)
 {
     int ans = n;
     for(int i=;i*i<=n;i++)
     {
         if(n%i == )
         {
             ans -= ans/i;
             while(n%i == ) n/= i;
         }
     }
     if(n>) ans -= ans/n;
     return ans;
 }
 int pow_mod(int x,int cnt)
 {
     int base = x%MOD,res = ;
     while(cnt)
     {
         if(cnt&) {res *= base;res %= MOD;}
         base *= base;base %= MOD;
         cnt >>= ;
     }
     return res;
 }

 int polya(int n,int m)
 {
     int res = ;
     int i;
     for(i=;i*i<n;i++)
     {
         if(n%i == )
         {
             res += Eular(i)%MOD * pow_mod(m,n/i-)%MOD;res %= MOD;
             res += Eular(n/i)%MOD * pow_mod(m,i-)%MOD;res %= MOD;
         }
     }
     if(i*i == n) res += Eular(i)%MOD*pow_mod(m,i-)%MOD;
     return res%MOD;
 }

 int main()
 {
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d",&N,&MOD);
         printf("%d\n",polya(N,N));
     }
 }