zoj 2317 Nice Patterns Strike Back(矩阵乘法)

http://acm.zju.edu.cn/onlinejudge/showProblem.do?

problemId=1317

给出一个n*m的矩阵(n <= 10^100, m <= 5)，对于2*2的子方格若全是黑色或全是白色的是非法的，用黑白两色去染n*m的方格，问共同拥有多少种合法的染色方案。

构造出转移矩阵，上一行向下一行的转移矩阵。由于m<=5，每行最多有32个状态。能够进行状态压缩构造出一个32*32的转移矩阵A。A[i][j] = 1表示上一行i状态能够向下一行的j状态转移。否则不能转移。要求最后的合法方案数，就再构造一个B矩阵，是一个32*1的矩阵，表示了到达第一行每个状态的方案数。那么A*B就表示到达第二行每个状态的方案数。以此类推。A^n-1 * B表示到达第n行每个状态的合法方案数，那么全部状态相应方案数的和就是总的方案数。

由于n特别大。要用到大数。我存矩阵的时候開始定义的大数类，一直T。改成了int型才A，1s+，难道大数类这么慢么，5s都过不了。

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();
}
}
}