标签：style blog color io os 使用 ar for sp
从左边开始染色，到第$i$个方块为止，红绿都是偶数的方案数为$a_i$，红绿恰有一个是偶数的方案为$b_i$，红绿都是奇数的方案为$c_i$，从而有如下状态转移方程：
$a_{i+1} = 2 \times a_i + b_i$
$b_{i+1} = 2 \times a_i + 2 \times b_i + 2 \times c_i$
$c_{i+1} = b_i + 2 \times c_i$
从而可以用矩阵表示：
\begin{equation} \left( \begin{array}{c} a_i \\ b_i \\ c_i \end{array} \right) = \left( \begin{array}{ccc} 2 & 1 & 0 \\ 2 & 2 & 2 \\ 0 & 1 & 2 \end{array} \right)^i \left( \begin{array}{c} 1\\ 0\\ 0\\ \end{array} \right) \end{equation}
最终使用矩阵的快速幂即可求解。
1 #define MOD 10007 2 #define M 3 3 #include <cstdio> 4 #include <cstdlib> 5 #include <iostream> 6 #include <cstring> 7 int N; 8 using namespace std; 9 class Matrix 10 { 11 public: 12 int a[M][M]; 13 Matrix(bool init = 0) 14 { 15 memset(a, 0, sizeof(a)); 16 if( init == 1 ) 17 { 18 for( int i = 0 ; i < M ; i++ ) 19 { 20 a[i][i] = 1; 21 } 22 } 23 } 24 int * operator [](int i) 25 { 26 return a[i]; 27 } 28 const int * operator [](int i)const 29 { 30 return a[i]; 31 } 32 }; 33 Matrix A; 34 void add(int &a, int b) 35 { 36 a += b; 37 while( a >= MOD ) 38 { 39 a -= MOD; 40 } 41 } 42 Matrix operator * (const Matrix &a, const Matrix &b) 43 { 44 Matrix c; 45 for( int i = 0 ; i < M ; i++ ) 46 { 47 for( int j = 0 ; j < M ; j++ ) 48 { 49 for( int k = 0 ; k < M ; k++ ) 50 { 51 add(c[i][k], 1ll * a[i][j] * b[j][k] %MOD); 52 } 53 } 54 } 55 return c; 56 } 57 Matrix quick_pow(Matrix m, int n) 58 { 59 Matrix ans(1); 60 while( n ) 61 { 62 if( n & 1 ) 63 { 64 ans = ans * m; 65 } 66 n = n>>1; 67 m = m * m; 68 } 69 return ans; 70 } 71 void solve() 72 { 73 Matrix ans; 74 ans = quick_pow(A, N); 75 printf ( "%d\n", ans[0][0]); 76 } 77 int main(int argc, char *argv[]) 78 { 79 int T; 80 scanf ( "%d", &T ); 81 A[0][0] = 2, A[0][1] = 1; 82 A[1][0] = 2, A[1][1] = 2, A[1][2] = 2; 83 A[2][1] = 1, A[2][2] = 2; 84 while( T-- ) 85 { 86 scanf ( "%d", &N ); 87 solve(); 88 } 89 }
标签：style blog color io os 使用 ar for sp
原文地址：http://www.cnblogs.com/jostree/p/4000809.html
