problemId=1317">http://acm.zju.edu.cn/onlinejudge/showProblem.do?

problemId=1317

給出一個n*m的矩陣(n <= 10^100, m <= 5)，對於2*2的子方格若全是黑色或全是白色的是非法的，用黑白兩色去染n*m的方格，問共同擁有多少種合法的染色方案。

構造出轉移矩陣，上一行向下一行的轉移矩陣。由於m<=5，每行最多有32個狀態。能夠進行狀態壓縮構造出一個32*32的轉移矩陣A。A[i][j] = 1表示上一行i狀態能夠向下一行的j狀態轉移。否則不能轉移。要求最後的合法方案數，就再構造一個B矩陣，是一個32*1的矩陣，表示了到達第一行每個狀態的方案數。那麽A*B就表示到達第二行每個狀態的方案數。以此類推。A^n-1 * B表示到達第n行每個狀態的合法方案數，那麽全部狀態相應方案數的和就是總的方案數。

由於n特別大。要用到大數。我存矩陣的時候開始定義的大數類，一直T。改成了int型才A，1s+，難道大數類這麽慢麽，5s都過不了。

import java.io.*;
import java.math.BigInteger;
import java.util.Scanner;

class Matrix
{
	public int [][] mat = new int[35][35];
	public Matrix()
	{
		
	}
	public void init()
	{
		for(int i = 0; i < 35; i++)
		{
			for(int j = 0; j < 35; j++)
			{
				if(i == j)
					mat[i][j] = 1;
				else mat[i][j] = 0;
			}
		}
	}
}

public class Main {
	public static Matrix A,B,res;
	public static BigInteger n;
	public static int m,p,test,mm;
	static Scanner cin = new Scanner(System.in);

	public static void buildA()
	{
		for(int i = 0; i < mm; i++)
		{
			for(int j = 0; j < mm; j++)
			{
				String s1 = Integer.toBinaryString(i);
				String s2 = Integer.toBinaryString(j);
				String ss1 = new String();
				String ss2 = new String();
				for(int k = 0; k < m-s1.length(); k++)
					ss1 += '0';
				ss1 += s1;
				for(int k = 0; k < m-s2.length(); k++)
					ss2 += '0';
				ss2 += s2;
				
				int flag = 0;
				for(int k = 0; k < m-1; k++)
				{
					if(ss1.charAt(k)-'0' == 0 && ss1.charAt(k+1)-'0' == 0 
					&& ss2.charAt(k)-'0' == 0 && ss2.charAt(k+1)-'0' == 0)
					{flag = 1;break;}
					if(ss1.charAt(k)-'0' == 1 && ss1.charAt(k+1)-'0' == 1
					&& ss2.charAt(k)-'0' == 1 && ss2.charAt(k+1)-'0' == 1)
					{flag = 1;break;}
				}
				if(flag == 1)
					A.mat[i][j] = 0;
				else A.mat[i][j] = 1;
			}
		}
	}
	
	
	public static Matrix Mul(Matrix x,Matrix y)
	{
		Matrix ans = new Matrix();
		for(int i = 0; i < mm; i++)
		{
			for(int k = 0; k < mm; k++)
			{
				if(x.mat[i][k] == 0)
					continue;
				for(int j = 0; j < mm; j++)
				{
					ans.mat[i][j] = (ans.mat[i][j]+x.mat[i][k]*y.mat[k][j])%p;
				}
			}
		}
		return ans;
	}
	
	public static Matrix Pow(Matrix tmp,BigInteger nn)
	{
		Matrix ans = new Matrix();
		ans.init();
		while(nn.compareTo(BigInteger.ZERO) > 0)
		{
			if( nn.remainder(BigInteger.valueOf(2)).compareTo(BigInteger.ONE) == 0)
				ans = Mul(ans,tmp);
			nn = nn.divide(BigInteger.valueOf(2));
			tmp = Mul(tmp,tmp);
		}
		return ans;
	}
	
	public static void main(String[] args) throws IOException 
	{
		int test = cin.nextInt();
		while((test--) > 0)
		{
			n = cin.nextBigInteger();
			m = cin.nextInt();
			p = cin.nextInt();
			mm = (1<<m);
			
			A = new Matrix();	
			B = new Matrix();
		
			buildA();
			for(int i = 0; i < mm; i++)
				B.mat[i][0] = 1;
						
			res = Pow(A,n.subtract(BigInteger.ONE));
			res = Mul(res,B);
			
			int anw = 0;
			for(int i = 0; i < mm; i++)
			{
				anw = (anw+res.mat[i][0])%p;
			}
			System.out.println(anw);
			if(test > 0)
				System.out.println();
		}
	}
}

zoj 2317 Nice Patterns Strike Back(矩陣乘法)