1 #include <iostream> 2 #include <cstdlib> 3 #include <cstring> 4 #include <queue> 5 #include <cstdio> 6 #include <algorithm> 7 #include <map> 8 #include <time.h> 9 #include <ext/pb_ds/assoc_container.hpp> 10 #include <ext/pb_ds/tree_policy.hpp> 11 #define LL long long 12 13 using namespace std; 14 using namespace __gnu_pbds; 15 16 //U[n]表示3×N棋盘的摆放数， 17 // 18 //V[n]表示缺了1×1边角的3×N棋盘的摆放数 19 // 20 //则 21 // 22 //U[n] = 2*V[n-1] + U[n-2] 23 // 24 //V[n] = U[n-1] + V[n-2] 25 // 26 //其中， 27 // 28 //U[0] = 1, U[2] = 3; 29 // 30 //V[0] = 1, V[1] =1; 31 32 //[U(n) [ 0 2 1 0 [ U(n-1) 33 // V(n) 1 0 0 1 V(n-1) 34 // U(n-1) = 1 0 0 0 * U(n-2) 35 // V(n-1)] 0 1 0 0 ] V(n-2) ] 36 // 37 // [ 0 2 1 0 [ U(1) 38 // = 1 0 0 1 V(1) 39 // 1 0 0 0 ^ (n-1) U(0) 40 // 0 1 0 0 ] V(0) ] 41 42 const int MOD = 12357; 43 44 struct Martix 45 { 46 LL martix[5][5]; 47 int row,col; 48 Martix(int _row,int _col) 49 { 50 memset(martix,0,sizeof(martix)); 51 row = _row; 52 col = _col; 53 } 54 Martix operator *(const Martix &A)const 55 { 56 Martix C(row, A.col); 57 for(int i = 0; i < row; i++) 58 for(int j = 0; j < A.col; j++) 59 for(int k = 0; k < col; k++) 60 C.martix[i][j] = (C.martix[i][j] + martix[i][k] * A.martix[k][j])%MOD; 61 return C; 62 } 63 }; 64 65 66 void solve() 67 { 68 Martix A(4,4), F(4,4), S(4,1); 69 A.martix[0][2] = A.martix[1][0] = A.martix[1][3] = A.martix[2][0] = A.martix[3][1] = 1; 70 F.martix[0][0] = F.martix[1][1] = F.martix[2][2] = F.martix[3][3] = 1; 71 A.martix[0][1] = 2; 72 S.martix[1][0] = S.martix[2][0] = S.martix[3][0] = 1; 73 int n; 74 scanf("%d",&n); 75 if(n&1) 76 { 77 printf("%d ",0); 78 return ; 79 } 80 n--; 81 while(n > 0) 82 { 83 if(n & 1) 84 F = F*A; 85 A = A*A; 86 n >>= 1; 87 } 88 S = F*S; 89 printf("%lld ",S.martix[0][0]); 90 } 91 92 int main(void) 93 { 94 solve(); 95 return 0; 96 }