POJ3420 Quad Tiling【矩阵快速幂】

Quad Tiling
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 5008 Accepted: 2269

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 10000
3 10000
5 10000
0 0

Sample Output
1
11
95

Source

POJ Monthly–2007.10.06, Dagger

问题链接：POJ3420 Quad Tiling
问题简述：（略）
问题分析：
递推式如下：
a[0]=1，a[1]=1，a[2]=5，a[3]=11，
a[n]=a[n-1]+5a[n-2]+a[n-3]-a[n-4](n >= 4)
采用矩阵快速模幂计算，模板题。关键是找到那个递推式。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* POJ3420 Quad Tiling */#include<cstdio>
#include<cstring>using namespace std;const int N = 10;
const int M = 10;int r, mod;struct Matrix
{int n,m;int a[N][M];void clear(){n = m = 0;memset(a, 0, sizeof(a));}Matrix operator +(const Matrix &b) const{Matrix c;c.n = n;c.m = m;for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)c.a[i][j] = a[i][j] + b.a[i][j];return c;}Matrix operator -(const Matrix &b) const{Matrix c;c.n = n;c.m = m;for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)c.a[i][j] = a[i][j] - b.a[i][j];return c;}Matrix operator *(const Matrix &b) const{Matrix c;c.clear();c.n = n;c.m = b.m;for(int i = 0; i < n; i++)for(int j = 0; j < b.m; j++)for(int k = 0; k < m; k++){c.a[i][j] += a[i][k] * b.a[k][j];c.a[i][j] %= mod;}return c;}Matrix powermod(int x){Matrix c;c.clear();c.n = c.m = n;for(int i = 0; i < n; i++)c.a[i][i] = 1;if(x == 0)return c;else if(x == 1)return *this;Matrix d = powermod(x / 2);d = d * d;if(x % 2 )d = d * (*this);return d;}
};int main()
{while(scanf("%d%d", &r, &mod) == 2 && (r || mod)) {int a[] = {1, 1, 5, 11};if(r <= 3) {printf("%d\n", a[r] % mod);continue;}Matrix b;b.clear();b.n = b.m = 4;b.a[0][1] = b.a[1][2] = b.a[2][3] = 1;b.a[3][0] = -1;b.a[3][1] = 1;b.a[3][2] = 5;b.a[3][3] = 1;b = b.powermod(r - 3);Matrix x;x.clear();x.n = 4;x.m = 1;x.a[0][0] = 1;x.a[1][0] = 1;x.a[2][0] = 5;x.a[3][0] = 11;x = b * x;printf("%d\n",(x.a[3][0] + mod) % mod);}return 0;
}

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POJ3420 Quad Tiling【矩阵快速幂】

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