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)
 1 /*--------------------------------------------------------------------------------------*/
 2 
 3 #include <algorithm>
 4 #include <iostream>
 5 #include <cstring>
 6 #include <ctype.h>
 7 #include <cstdlib>
 8 #include <cstdio>
 9 #include <vector>
10 #include <string>
11 #include <queue>
12 #include <stack>
13 #include <cmath>
14 #include <set>
15 #include <map>
16 
17 //debug function for a N*M array
18 #define debug_map(N,M,G) printf("
");for(int i=0;i<(N);i++)
19 {for(int j=0;j<(M);j++){
20 printf("%d",G[i][j]);}printf("
");}
21 //debug function for int,float,double,etc.
22 #define debug_var(X) cout<<#X"="<<X<<endl;
23 #define LL long long
24 const int INF = 0x3f3f3f3f;
25 const LL LLINF = 0x3f3f3f3f3f3f3f3f;
26 /*--------------------------------------------------------------------------------------*/
27 using namespace std;
28 
29 int N,M,T;
30 int MOD = 1e9+7;
31 
32 int Eular(int n)
33 {
34     int ans = n;
35     for(int i=2;i*i<=n;i++)
36     {
37         if(n%i == 0)
38         {
39             ans -= ans/i;
40             while(n%i == 0) n/= i;
41         }
42     }
43     if(n>1) ans -= ans/n;
44     return ans;
45 }
46 int pow_mod(int x,int cnt)
47 {
48     int base = x%MOD,res = 1;
49     while(cnt)
50     {
51         if(cnt&1) {res *= base;res %= MOD;}
52         base *= base;base %= MOD;
53         cnt >>= 1;
54     }
55     return res;
56 }
57 
58 int polya(int n,int m)
59 {
60     int res = 0;
61     int i;
62     for(i=1;i*i<n;i++)
63     {
64         if(n%i == 0)
65         {
66             res += Eular(i)%MOD * pow_mod(m,n/i-1)%MOD;res %= MOD;
67             res += Eular(n/i)%MOD * pow_mod(m,i-1)%MOD;res %= MOD;
68         }
69     }
70     if(i*i == n) res += Eular(i)%MOD*pow_mod(m,i-1)%MOD;
71     return res%MOD;
72 }
73 
74 int main()
75 {
76     scanf("%d",&T);
77     while(T--)
78     {
79         scanf("%d%d",&N,&MOD);
80         printf("%d
",polya(N,N));
81     }
82 }