POJ3420_Quad Tiling_矩阵与递推

Quad Tiling

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4363 Accepted: 1996

Description

Tired of the Tri Tiling game finally, Michael turns to a more challengeable game, Quad Tiling:

In how many ways can you tile a 4 × N (1 ≤ N ≤ 10⁹) rectangle with 2 × 1 dominoes? For the answer would be very big, output the answer modulo M (0 < M ≤ 10⁵).

Input

Input consists of several test cases followed by a line containing double 0. Each test case consists of two integers, N and M, respectively.

Output

For each test case, output the answer modules M.

Sample Input

1 100003 100005 100000 0

Sample Output

11195

用 1 * 2 的砖铺满 4 * n 的地面，其中砖旋转 90 度，求有多少种铺法。

用递推，观察铺满第 i 列之后，i+1 列总共有 16 种状态。我们要求的是铺满第 n 列后，n + 1 列为空的状态。

因此，从为空的这一状态出发，求状态转移的递推式子。最后发现只有 6 种情况参与了递推，其中有两种还可以合并。由此写处了 5 阶转移方阵。

起始时，铺满第 0 列后第 1 列的状态为空，所以为空的种数是 1， 其他都是 0.

最后用矩阵快速幂进行变换就好了。

#include<cstdio>#include<iostream>#include<vector>using namespace std;typedef vector<int> vec;typedef vector<vec> mat;int n, m;mat Mul(mat A, mat B){mat C(A.size(), vec(B[0].size()));for(int i= 0; i< A.size(); i++)for(int k= 0; k< B.size(); k++)for(int j= 0; j< B[0].size(); j++)C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % m;return C;}mat Pow(mat A, int n){mat B(A.size(), vec(A.size()));for(int i= 0; i< A.size(); i++)B[i][i] = 1;while(n){if(n & 1) B = Mul(B, A);A = Mul(A, A);n = n >> 1;}return B;}int main(){mat A(5, vec(1));A[0][0] = 1;mat B(5, vec(5));B[0][0] = B[0][1] = B[0][2] = B[0][3] = 1;B[1][0] = 1;B[2][0] = B[2][4] = 1;B[3][0] = 2, B[3][3] = 1;B[4][2] = 1;while(scanf("%d %d", &n, &m), m){mat res(5, vec(1));res = Mul(Pow(B, n), A);printf("%d\n", res[0][0]);}return 0;}