bzoj 4330: JSOI2012 爱之项链

听说这题不公开.. 那就不贴题意了

首先要用burnside引理求出戒指的种数，那么对于一个顺时针旋转$k$个位置的置换就相当于连上一条$(i,(i+k)%R)$的边，每个环颜色必须相同

环的个数为$gcd(k,R)$，所以这样的方案数就有$R^{gcd(k,R)}$种

然后dp求项链的方案数，设$g_{i,0}$表示$1$和$i$不同，中间相邻不同的方案数，$g_{i,1}$表示$1$和$i$相同，中间相邻不同的方案数

那么有如下推导：

$$g_{i,0}=(i-1)g_{i-1,1}+(i-2)g_{i-1,0}$$

$$g_{i,1}=g_{i-1,0}$$

再化一下：$$g_{i,0}=(i-1)g_{i-2,0}+(i-2)g_{i-1,0}$$

由于我们需要的也就是$g_{i,0}$，不妨设$f_i=g_{i,0}$，所以：$$f_i=(i-1)f_{i-2}+(i-2)f_{i-1}$$

这就可以用矩阵乘法优化 听说有通项公式

 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>
 #include <cmath>
 #define LL long long
 using namespace std;
 const LL Mod = ;
 const LL Maxn = ;
 LL n, m, r;
 LL gcd ( LL a, LL b ){
     if ( a ==  ) return b;
     return gcd ( b%a, a );
 }
 LL pow ( LL x, LL k ){
     LL ret = ;
     while (k){
         if ( k &  ) ret = (ret*x)%Mod;
         x = (x*x)%Mod;
         k >>= ;
     }
     return ret;
 }
 struct matrix {
     LL a[][];
     LL l1, l2;
     void clear (){ memset ( a, , sizeof (a) ); }
 }trans, x, z, fi;
 matrix ttimes ( matrix x, matrix y ){
     matrix ret;
     ret.clear ();
     ret.l1 = x.l1; ret.l2 = y.l2;
     LL i, j, k;
     for ( i = ; i < ret.l1; i ++ ){
         for ( j = ; j < ret.l2; j ++ ){
             for ( k = ; k < x.l2; k ++ ){
                 ret.a[i][j] = ( ret.a[i][j] + (x.a[i][k]*y.a[k][j])%Mod ) % Mod;
             }
         }
     }
     return ret;
 }
 LL phi ( LL x ){
     LL ret = x, sq = sqrt (x);
     for ( LL i = ; i <= sq; i ++ ){
         if ( x % i ==  ){
             ret = ret/i*(i-);
             while ( x % i ==  ) x /= i;
         }
     }
     if ( x >  ) ret = ret/x*(x-);
     return ret;
 }
 int main (){
     LL i, j, k;
     scanf ( "%lld%lld%lld", &n, &m, &r );
     LL o = , sq = sqrt (m);
     for ( i = ; i <= sq; i ++ ){
         if ( m % i ==  ){
             o = (o+(pow(r,i)*phi(m/i))%Mod)%Mod;
             if ( i*i == m ) continue;
             o = (o+(pow(r,m/i)*phi(i))%Mod)%Mod;
         }
     }
     LL inv = pow ( m, Mod- );
     o = (o*inv)%Mod;
     if ( n ==  ){ printf ( "0\n" ); return ; }
     trans.l1 = trans.l2 = ;
     trans.a[][] = o-; trans.a[][] = ; trans.a[][] = o-;
     x = trans;
     z.l1 = z.l2 = ;
     z.a[][] = ; z.a[][] = ;
     for ( i = n-; i >= ; i >>=  ){
         if ( i &  ) z = ttimes ( z, x );
         x = ttimes ( x, x );
     }
     fi.l1 = ; fi.l2 = ;
     fi.a[][] = (o*(o-))%Mod;
     fi = ttimes ( fi, z );
     printf ( "%lld\n", fi.a[][] );
     return ;
 }