ZOJ2317-Nice Patterns Strike Back：矩陣快速冪，高精度

標籤：des   style   blog   http   color   io   os   ar   java   

Nice Patterns Strike BackTime Limit: 20000/10000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others)Problem Description

      You might have noticed that there is the new fashion among rich people to have their yards tiled with black and white tiles, forming a pattern. The company Broken Tiles is well known as the best tiling company in our region. It provides the widest choices of nice patterns to tile your yard with. The pattern is nice if there is no square of size 2 × 2, such that all tiles in it have the same color. So patterns on the figure 1 are nice, while patterns on the figure 2 are not.

      The president of the company wonders whether the variety of nice patterns he can provide to the clients is large enough. Thus he asks you to find out the number of nice patterns that can be used to tile the yard of size N × M . Now he is interested in the long term estimation, so he suggests N ≤ 10100. However, he does not like big numbers, so he asks you to find the answer modulo P .

Input      The input file contains three integer numbers: N (1 ≤ N ≤ 10100), M (1 ≤ M ≤ 5) and P (1 ≤ P ≤10000).Output      Write the number of nice patterns of size N × M modulo P to the output file.Sample Input

2 2 53 3 23

Sample Output

40

SourceAndrew Stankevich Contest 1  演算法：因為m<=5，每一行的狀態可以用一個位元表示，構造係數矩陣A，其中aij為1表示狀態i和j不衝突，為0表示衝突，結果為A^(n-1)中矩陣各元素之和，由於n很大，所以涉及到高精度和矩陣快速冪，所以我用Java寫了。   

  1 import java.awt.Checkbox;  2 import java.io.BufferedInputStream;  3 import java.io.BufferedOutputStream;  4 import java.io.PrintWriter;  5 import java.math.BigInteger;  6 import java.util.Scanner;  7   8 public class Main {  9  10     static int p; 11  12     public static class Matrix implements Cloneable { 13         long[][] a; 14         int d; 15  16         public Matrix(int d) { 17             this.d = d; 18             a = new long[d][d]; 19         } 20  21         public Matrix multiply(Matrix m) { 22             Matrix ret = new Matrix(d); 23             for (int i = 0; i < d; ++i) { 24                 for (int j = 0; j < d; ++j) { 25                     for (int k = 0; k < d; ++k) { 26                         ret.a[i][j] += a[i][k] * m.a[k][j]; 27                         ret.a[i][j] %= p; 28                     } 29                 } 30             } 31             return ret; 32         } 33  34         public Matrix clone() { 35             Matrix ret = new Matrix(d); 36             ret.a = a.clone(); 37             return ret; 38         } 39  40         Matrix pow(BigInteger cnt) { 41             // 先產生一個單位矩陣 42             Matrix eye = new Matrix(d); 43             for (int i = 0; i < d; i++) 44                 eye.a[i][i] = 1; 45  46             for (int i = cnt.bitLength() - 1; i >= 0; i--) { 47                 eye = eye.multiply(eye); 48                 if (cnt.testBit(i)) { 49                     eye = eye.multiply(this); 50                 } 51             } 52             return eye; 53         } 54     } 55  56     static boolean check(int x, int y, int m) { 57         for (int i = 1; i < m; i++) { 58             if ((x & 3) == (y & 3) && (x & 1) == ((x & 2) >> 1)) { 59                 return false; 60             } 61             x >>= 1; 62             y >>= 1; 63         } 64  65         return true; 66     } 67  68     public static void main(String[] args) { 69  70         Scanner cin = new Scanner(new BufferedInputStream(System.in)); 71         PrintWriter cout = new PrintWriter(new BufferedOutputStream(System.out)); 72  73         int T = cin.nextInt(); 74  75         while (T-- != 0) { 76             BigInteger n = cin.nextBigInteger(); 77             int m = cin.nextInt(); 78             p = cin.nextInt(); 79  80             // 產生矩陣A 81             int d = (1 << m); 82             Matrix A = new Matrix(d); 83             for (int i = 0; i < d; i++) 84                 for (int j = 0; j < d; j++) { 85                     if (check(i, j, m)) 86                         A.a[i][j] = 1; 87                 } 88  89             A = A.pow(n.subtract(BigInteger.ONE)); 90  91             long ans = 0; 92             for (int i = 0; i < d; i++) 93                 for (int j = 0; j < d; j++) { 94                     ans = (ans + A.a[i][j]) % p; 95                 } 96  97             cout.println(ans); 98             if (T != 0) 99                 cout.println("");100             // System.out.println(ans);101         }102 103         cin.close();104         cout.close();105 106     }107 }

 

ZOJ2317-Nice Patterns Strike Back：矩陣快速冪，高精度