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
题意：题面很臭很长。大意是，有一个大小为N的环，给出M，K，D，以及N个数。我们进行K次操作，每次操作把距离当前点不超过D的累加到当前点，结果模M。
思路：因为要进行K次，每次的原则是一样的，我们可以想到用矩阵来优化，如果i能到达j，把么base[i][j]=1；则结果ans=A*(base^K)。
但是需要优化，时间复杂度为O(N^3*lgK)。我们发现矩阵是下一行由上一行右移一位而来，那么我们保存一维即可代表这个矩阵。同样的，我们只需要得到第一行的矩阵结果，就能得到整个矩阵的结果。
#include<bits/stdc++.h>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=;
int N,Mod,D,K;
struct mat
{
int M[maxn];
mat(){ rep(i,,N) M[i]=; }
mat friend operator*(mat a,mat b){
mat res;
rep(k,,N)
rep(j,,N){
res.M[j]=(res.M[j]+(ll)a.M[k]*b.M[j-k+>?j-k+:j-k++N]%Mod)%Mod;
}
return res;
}
mat friend operator ^(mat a,int x)
{
mat res;rep(i,,N) res.M[]=;
while(x){
if(x&) res=res*a; a=a*a; x/=;
} return res;
} };
int main()
{
scanf("%d%d%d%d",&N,&Mod,&D,&K);
mat a,base;
rep(i,,N) scanf("%d",&a.M[i]);
rep(i,,N)
if(i-<=D||N-i+<=D||N-+i<=D) base.M[i]=;
a=a*(base^K);
rep(i,,N-) printf("%d ",a.M[i]);
printf("%d\n",a.M[N]);
return ;
}