POJ 3150 Cellular Automaton（矩阵快速幂）

Cellular Automaton

Time Limit: 12000MS Memory Limit: 65536K

Total Submissions: 3504 Accepted: 1421

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. 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

这道题目的矩阵好找，但是由于n比较大，用n*n的矩阵再加上快速幂，是O(n^3*log k) 回超时。观察矩阵，发现矩阵是一个循环矩阵，无论矩阵取多少次方，矩阵的每一行相当于第一行向后推了一步，所以说是循环矩阵，这样我们只要计算矩阵的第一行就可以知道矩阵的其他行，所以只开一维数组效率就是O(n^2log k)

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
typedef long long int LL;
int n,m,d,k;
struct Node
{
    LL a[505];
};
Node multiply(Node a,Node b)
{
    Node c;
    memset(c.a,0,sizeof(c.a));
    for(int i=0;i<n;i++)
    {
        int cnt=(n-i)%n;
        for(int j=0;j<n;j++)
        {
            (c.a[i]+=(a.a[j]*b.a[cnt++])%m)%=m;
            if(cnt==n) cnt=0;
        }
    }
    return c;
}
Node get(Node a,int x)
{
    Node c;
    memset(c.a,0,sizeof(c.a));
    c.a[0]=1;
    for(x;x;x>>=1)
    {
        if(x&1) c=multiply(c,a);
        a=multiply(a,a);
    }
    return c;
}
int main()
{
    scanf("%d%d%d%d",&n,&m,&d,&k);
    Node a;Node b;
    memset(a.a,0,sizeof(a.a));
    memset(b.a,0,sizeof(b.a));
    for(int i=0;i<n;i++)
       scanf("%lld",&b.a[i]);
    a.a[0]=1;
    for(int i=1;i<=d;i++)
      a.a[i]=a.a[n-i]=1;
    a=get(a,k);
    a=multiply(b,a);
    for(int i=0;i<n;i++)
       if(i==n-1) printf("%lld\n",a.a[i]);
       else printf("%lld ",a.a[i]);
    return 0;
}