POJ 3420Quad Tiling

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

111
95
由于宽度只有4，所以我们可以用[0,15]这16个数字的二进制形式来表示每一行的状态，
然后考虑状态的转移情况，这一行的状态从前一行的那些状态转移而来。
这样我们可以得到一个16*16的矩阵，然后给出初始的1*16的矩阵。
接下来就是矩阵快速幂了。
#include<set>#include<map>#include<ctime>#include<cmath>#include<stack>#include<queue>#include<bitset>#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>#include<functional>#define rep(i,j,k) for (int i = j; i <= k; i++)#define per(i,j,k) for (int i = j; i >= k; i--)using namespace std;typedef long long LL;const int low(int x) { return x&-x; }const double eps = 1e-8;const int INF = 0x7FFFFFFF;//const int mod = 1e9 + 7;const int N = 15;int T, n, mod;int a[N + 1][N + 1], b[N + 1][N + 1], c[N + 1][N + 1], ans[N + 1];int main(){rep(i, 0, N) a[i][i ^ N] = 1;rep(i, 0, N){if ((i & 3) == 3) a[N ^ i ^ 3][i] = 1;if ((i & 6) == 6) a[N ^ i ^ 6][i] = 1;if ((i & 12) == 12) a[N ^ i ^ 12][i] = 1;if (i == N) a[i][i] = 1;}while (scanf("%d%d", &n, &mod) != EOF, n&&mod){rep(i, 0, N) rep(j, 0, N) b[i][j] = a[i][j];memset(ans, 0, sizeof(ans));ans[N] = 1;for (; n; n >>= 1){if (n & 1){rep(i, 0, N){c[0][i] = 0;rep(j, 0, N){c[0][i] = (c[0][i] + 1LL * ans[j] * b[j][i]) % mod;}}rep(i, 0, N) ans[i] = c[0][i];}rep(i, 0, N) rep(j, 0, N){c[i][j] = 0;rep(k, 0, N){c[i][j] = (c[i][j] + 1LL * b[i][k] * b[k][j]) % mod;}}rep(i, 0, N) rep(j, 0, N) b[i][j] = c[i][j]; }printf("%d\n", ans[N]);}return 0;}