[POJ 3150][矩阵快速幂] [循环矩阵]Cellular Automaton

题目描述：
题目链接：POJ 3150 Cellular Automaton
题面：
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(|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.

题目大意：
n个数围成一个环，进行k次操作。每次操作将所有数进行变换，对于第i个数：与第i个数距离小于等于d的数（包括i）加起来的和对m取模的值，作为新的第i个数。求最后k次操作后，每个位置的数是多少。下图为n=5,m=3,d=1,k=1：

输入输出格式：
多组数据，每组数据第一行为n, m, d, k (1 ≤ n ≤ 500, 1 ≤ m ≤ 1 000 000, 0 ≤ d < $\frac{n}{2}$ , 1 ≤ k ≤ 10 000 000)。第二行为n个数（ ≤ m-1）。
每组数据输出一行，即k次操作后，n个数分别是多少。
样例输入输出：
Input

5 3 1 1
1 2 2 1 2
5 3 1 10
1 2 2 1 2

Output

2 2 2 2 1
2 0 0 2 2

题目分析：
首先我们很容易就能发现可以利用矩阵来求，因为根据题意求每个数，其实就是与其距离小于等于d的几个数的和。以样例为例，A= 1 2 2 1 2
B =
1 1 0 0 1
1 1 1 0 0
0 1 1 1 0
0 0 1 1 1
1 0 0 1 1
操作一次，就是 $A\ast B$的结果。
所以答案就是 $A\ast B^k$。
于是我就直接写了矩阵快速幂交上去，然后TLE。因为对于 n 2 n^2