Time Limit: 12000MS | Memory Limit: 65536K | |
Total Submissions: 3048 | Accepted: 1227 | |
Case Time Limit: 2000MS |
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. Theorder 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 orderm. 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
/* ID: wuqi9395@126.com PROG: LANG: C++ */ #include<map> #include<set> #include<queue> #include<stack> #include<cmath> #include<cstdio> #include<vector> #include<string> #include<fstream> #include<cstring> #include<ctype.h> #include<iostream> #include<algorithm> #define INF (1<<30) #define PI acos(-1.0) #define mem(a, b) memset(a, b, sizeof(a)) #define For(i, n) for (int i = 0; i < n; i++) typedef long long ll; using namespace std; const int maxn = 505; const int maxm = 505; int mod, n, k, d; void Matrix_pow(int a[], int b[]) { ll c[505] = {0}; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) c[i] += ((ll)a[j] * b[(i + j) % n]) % mod; c[i] %= mod; } for (int i = 0; i < n; i++) b[i] = c[i]; } int a[505], b[505]; int main () { while(scanf("%d%d%d%d", &n, &mod, &d, &k) != EOF) { for (int i = 0; i < n; i++) scanf("%d", b + i); memset(a, 0, sizeof(a)); a[0] = 1; for (int i = 1; i <= d; i++) a[i] = a[n - i] = 1; while(k) { if (k & 1) Matrix_pow(a, b); k >>= 1; Matrix_pow(a, a); } for (int i = 0; i < n; i++) printf("%d%c", b[i], " \n"[i == n - 1]); } return 0; }
[POJ 3150] Cellular Automaton (矩阵快速幂 + 矩阵乘法优化),布布扣,bubuko.com
[POJ 3150] Cellular Automaton (矩阵快速幂 + 矩阵乘法优化)
原文地址:http://blog.csdn.net/sio__five/article/details/37814569