poj 2888 Magic Bracelet

一道很不错的题目，这里加入连接限制同时还考察优化，优化方法同上。连接限制如何处理？注意到项链个数很少，因此可以建图，然后分别求出每种 颜色连接n个珠子后回到自身的方案数，累加即可，这里可以用矩阵快速幂求解。

 

View Code

#include<iostream>

 #include<cstdio>

 #include<cstdlib>

 #include<algorithm>

 #include<cmath>

 #include<queue>

 #include<set>

 #include<map>

 #include<cstring>

 #include<vector>

 #include<string>

 #define LL long long

 #define Mod 9973

 using namespace std;

 int prime[4500],cnt;

 class Matrix

 {

 public:

       int matrix[15][15];

       void Zero(  ){ memset( matrix ,0 , sizeof( matrix ) ); }

       int n;

       Matrix operator *( Matrix &b );    

 };

 void Prime(  )

 {

     bool hash[20024] = {0};

     for( int i = 3 ; i <= 200 ; i += 2 ){

             if( !hash[i>>1] ){

                 int x = i << 1;

                 for( int j = i*i; j <= 40000; j+=x )

                      hash[j>>1] = true;

                 }

        }    

     cnt = 0;

     prime[cnt++] = 2;

     for( int i = 1; i <= 20000; i ++ ){

         if( !hash[i] ) prime[cnt++] = i <<1|1;

         }

 }

 Matrix Matrix::operator *( Matrix &b )

 {

       Matrix t;

       t.Zero();

       for( int i = 0 ;i < n; i ++ ){

            for( int k = 0 ; k < n ; k ++ ){

                 if( matrix[i][k] ){

                     for( int j = 0 ; j < n ; j ++ ){

                          if( b.matrix[k][j] )

                              t.matrix[i][j] = ( t.matrix[i][j] + matrix[i][k]*b.matrix[k][j] )%Mod;

                     }    

                 }        

            }    

       }    

       t.n = n;    

       return t;

 }

 int Eular( int n ){

     int ans = n;

     for( int i = 0 ; i < cnt ; i ++ ){

         if( n < prime[i] ) break;

         if( n % prime[i] == 0 ){

             ans = ans/prime[i]*(prime[i]-1);

             while( n % prime[i] == 0 ) {

                 n /= prime[i];

                 }

             }

         }

         if( n != 1 ) ans = ans/n*(n-1);

         return ans%Mod;

 }

 int Pow( int c, int k ){

     int ans = 1;

     c %= Mod;

     while( k ){

         if( k&1 ) ans = (ans*c)%Mod;

         c = (c*c)%Mod ;

         k >>= 1;

         }

     return ans;

     }

 int Solve( Matrix A ,int n  ){

     Matrix t;

     t.Zero();

     for( int i = 0; i < A.n; i ++ ) t.matrix[i][i] = 1;

     t.n = A.n;

     while( n ){

         if( n & 1 ) t =t*A;

         A = A*A;

         n >>= 1;

         }

     int sum =0;

     for( int i = 0 ; i < A.n ; i ++ )

          sum = (sum +t.matrix[i][i])%Mod;

     return sum;

     }

 int Polya( Matrix A,int n ,int k ){

     int t = (int)sqrt( (double)n ),sum  = 0;

     for( int i = 1; i <= t; i ++  ){

         if( n % i == 0 ){

     //        printf( "%d %d %d %d %d\n",sum,Eular(i),Eular(n/i ),Solve( A,i ),Solve( A , n /i ) );

             if( i*i == n )

                sum = ( sum + Eular( i )*Solve(A, i ) )%Mod;

              else

                  sum = ( sum + Eular( i )*Solve( A,n/i )+ Eular( n/i )*Solve(A , i ) )%Mod;

             }

         }

     sum = (sum * Pow( n , Mod - 2 ))%Mod;

     return sum;

     }

 int main(  )

 {

     int n , m , T,k,a,b;

     Prime();

     int p = 

     scanf( "%d",&T );

     while( T-- ){

           scanf( "%d %d %d",&n,&m,&k );

           Matrix A;A.Zero();

           A.n = m;

           for( int i = 0 ; i < m ; i ++ ){

             for( int j = 0 ; j < m ; j ++ ){

                 A.matrix[i][j] = 1;

                 }

             }

         for( int i = 0 ; i < k ;i ++ ){

              scanf( "%d %d",&a,&b );

              A.matrix[a-1][b-1] = 0;

              A.matrix[b-1][a-1] = 0;

         }

         printf( "%d\n",Polya( A,n ,k ) );

     }  

     //system( "pause" );

     return 0;

 }