[POJ 3150] Cellular Automaton (矩阵快速幂 + 矩阵乘法优化）

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.

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

/*
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;
}