POJ3150 --- Cellular Automaton (matrix)

POJ3150 --- Cellular Automaton (matrix)

Description

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 states of neighboring cells. the order of the cellular automation 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 (| I? J |, n? | I? J |). 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.

Input

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 <n? 2, 1 ≤ k ≤ 10 000 000). The second line contains n integer numbers from 0 to m? 1-initial values of the automaton's cells.

Output

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

Sample Input

Sample input #1
5 3 1 1
1 2 2 1 2

Sample input #2
5 3 1 10
1 2 2 1 2

Sample Output

Sample output #1
2 2 2 2 1

Sample output #2
2 0 0 2 2

Source
Northeastern Europe 2006.

The transfer matrix is easy to introduce, but it is too large to work.
We can see that each row of this matrix is cyclical, And the next row is moving one to the right of the previous row. Therefore, we can reduce the matrix multiplication to n ^ 2 and save it with only one-dimensional arrays, in this way, we can combine the Matrix to quickly power the ac.

/*************************************** * *********************************> File Name: POJ3150.cpp> Author: ALex> Mail: zchao1995@gmail.com> Created Time: ******************************** **************************************** /# include# Include
   
    
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             # Include using namespace std; const double pi = acos (-1.0); const int inf = 0x3f3f3f; const double eps = 1e-15; typedef long LL; typedef pair
            
              PLL; LL mat [505] [505]; LL base [505]; LL arr [505]; LL tmp [505]; LL init [505]; int n, m, d, k; void mul (LL a []) {for (int j = 1; j <= n; ++ j) {tmp [j] = 0; for (int I = 1; I <= n; ++ I) {tmp [j] + = mat [I] [j] * a [I]; tmp [j] % = m ;}}for (int I = 1; I <= n; ++ I) {a [I] = tmp [I] ;}} void fastpow () {while (k) {if (k & 1) {mul (arr);} k >>=1; mul (base ); for (int I = 1; I <= n; ++ I) {for (int j = 1; j <= n; + + J) {int p = (j + I-1) % n; if (! P) {p = n;} mat [I] [p] = base [j] ;}}} int main () {while (~ Scanf ("% d", & n, & m, & d, & k) {for (int I = 1; I <= n; ++ I) {scanf ("% lld", & init [I]) ;}for (int j = 1; j <= n; ++ j) {for (int I = 1; I <= n; ++ I) {int dis = min (abs (I-j), n-abs (I-j )); if (dis <= d) {mat [I] [j] = 1 ;}} memset (arr, 0, sizeof (arr); arr [1] = 1; // unit matrix for (int I = 1; I <= n; ++ I) {base [I] = mat [1] [I];} fastpow (); for (int I = 1; I <= n; ++ I) {for (int j = 1; j <= N; ++ j) {int p = (j + I-1) % n; if (! P) {p = n;} mat [I] [p] = arr [j] ;}} mul (init); for (int I = 1; I <= n; ++ I) {printf ("% lld", init [I]); if (I <n) {printf ("");}} printf ("\ n");} return 0 ;}