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2013
11-12

POJ 3150 Cellular Automaton [解题报告] Java

Cellular Automaton

问题描述 :

A cellular automaton is a collection of cells on a grid of specified shape that evolves through a number of discrete time steps according to a set of rules that describe the new state of a cell based on the states of neighboring cells. The order of the cellular automaton is the number of cells it contains. Cells of the automaton of order n are numbered from 1 to n.

The order of the cell is the number of different values it may contain. Usually, values of a cell of order m are considered to be integer numbers from 0 to m − 1.

One of the most fundamental properties of a cellular automaton is the type of grid on which it is computed. In this problem we examine the special kind of cellular automaton — circular cellular automaton of order n with cells of order m. We will denote such kind of cellular automaton as n,m-automaton.

A distance between cells i and j in n,m-automaton is defined as min(|ij|, n − |ij|). A d-environment of a cell is the set of cells at a distance not greater than d.

On each d-step values of all cells are simultaneously replaced by new values. The new value of cell i after d-step is computed as a sum of values of cells belonging to the d-enviroment of the cell i modulo m.

The following picture shows 1-step of the 5,3-automaton.

The problem is to calculate the state of the n,m-automaton after k d-steps.

输入:

The first line of the input file contains four integer numbers n, m, d, and k (1 ≤ n ≤ 500, 1 ≤ m ≤ 1 000 000, 0 ≤ d < n2 , 1 ≤ k ≤ 10 000 000). The second line contains n integer numbers from 0 to m − 1 — initial values of the automaton’s cells.

输出:

Output the values of the n,m-automaton’s cells after k d-steps.

样例输入:

sample input #1
5 3 1 1
1 2 2 1 2

sample input #2
5 3 1 10
1 2 2 1 2

样例输出:

sample output #1
2 2 2 2 1

sample output #2
2 0 0 2 2

解题代码:

import java.util.Scanner;  
      
    public class Main{  
      
        static int n,m,d,k;  
          
          
        public static void main(String[] args) {  
              
            Scanner scan = new Scanner(System.in);  
              
            while(scan.hasNext()){  
                n = scan.nextInt();  
                m = scan.nextInt();  
                d = scan.nextInt();  
                k = scan.nextInt();  
                  
                long[] init = new long[n];  
                long[] tmp = new long[n];  
                  
                for(int i=0;i< n;i++)  
                    init[i] = scan.nextInt();  
                tmp[0] = 1;  
                for(int i=1;i<=d;i++)  
                    tmp[i] = tmp[n-i] = 1;  
                  
                  
                while(k!=0){  
                    if(k%2==1)  
                        mul(tmp,init);  
                    mul(tmp,tmp);  
                    k >>= 1;  
                }  
                  
                for(int i=0;i< n;i++)  
                    System.out.print(init[i]+" ");  
                System.out.println();  
                  
            }  
      
        }  
      
      
        public static void mul(long[] tmp, long[] init) {  
              
            long c[] = new long[n];  
              
            for(int i=0;i< n;i++){  
                for(int j=0;j< n;j++)  
				 //由矩阵第一行 可以通过 平衡获得所有列  
                    c[i] += tmp[j]*init[(i+j)%n]; 
            }  
              
            for(int i=0;i< n;i++)  
                init[i] = c[i]%m;  
              
              
        }  
      
    }

  1. 约瑟夫也用说这么长……很成熟的一个问题了,分治的方法解起来o(n)就可以了,有兴趣可以看看具体数学的第一章,关于约瑟夫问题推导出了一系列的结论,很漂亮

  2. 这个方法是错的,不信你试试:
    20 5
    1 A:9
    1 A:9
    1 A:9
    1 A:6
    1 A:4
    正确答案应该是19,这个答案是18