POJ3150---Cellular Automaton(矩阵) - c++编程基础

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 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.

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

容易推出转移矩阵,但是规模太大,还不行
观察发现这个矩阵的每一行都是循环的,下一行的就是上一行向右循环移动一位,所以我们可以把矩阵乘法降到n^2,只用一维数组来保存,这样再结合矩阵快速幂就可以ac了

/*************************************************************************
    > File Name: POJ3150.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年03月17日 星期二 20时42分20秒
 ************************************************************************/

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               using namespace std; const double pi = acos(-1.0); const int inf = 0x3f3f3f3f; const double eps = 1e-15; typedef long 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%d%d%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; //单位矩阵 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; }