POJ 3150 Cellular Automaton

Cellular Automaton

Time Limit: 12000MS Memory Limit: 65536KTotal Submissions: 3553 Accepted: 1434Case 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 #15 3 1 11 2 2 1 2sample input #25 3 1 101 2 2 1 2

Sample Output

sample output #12 2 2 2 1sample output #22 0 0 2 2

将初始值看做一个1*n的矩阵s，分析题意，容易得到ans=s* A^k;

（A是根据题意构造的矩阵）

时间复杂度O（N^3），显然爆炸。

仔细观察发现A是一个循环矩（方）阵。

循环方阵A满足：A[x][y]=A[1][(y-x+n)%n+1]； 

我们只计算和存储A的第一行，时空复杂度下降到O（n^2）可以通过。

代码如下：

#include<cstdio>#include<iostream>#include<cstdlib>#include<cmath>#include<cstring>#include<queue>#include<vector>#include<algorithm>#define LL long long#define CLEAR(xxx) memset(xxx,0,sizeof(xxx))using namespace std;const LL maxn=500+5,inf=1e9;LL n,m,d,k,s[maxn];LL A[maxn],C[maxn],ans[maxn],z[maxn];void multi(LL *A,LL *B){CLEAR(C);for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)(C[i]+=A[j]*B[(i-j+n)%n+1])%=m;memcpy(A,C,sizeof(C));} void Pow(int x){int i,k,j;CLEAR(ans); ans[1]=1;for(;x;x>>=1,multi(A,A))if(x&1)multi(ans,A);}void solve(){int i,j;CLEAR(A);CLEAR(s);for(i=1;i<=n;i++)scanf("%d",&s[i]);for(i=1;i<=d+1;i++)A[i]=1;for(i=n-d+1;i<=n;i++)A[i]=1;Pow(k);multi(s,ans);for(i=1;i<=n;i++)printf("%d ",C[i]);putchar('\n'); }int main(){while(cin>>n>>m>>d>>k)solve();return 0;}

<pre name="code" class="cpp">#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<cstdlib>#define mod m#define clear(xxx) memset(xxx,0,sizeof(xxx))using namespace std;const long long maxn=505;typedef long long arr[maxn];arr a,b,ans,z;long long n,m,d,k;void multiply(arr x,arr y){     memset(z,0,sizeof(z));     for(long long i=1;i<=n;i++){         for(long long h=1;h<=n;h++){             z[i]=(z[i]+x[h]*y[(i-h+n)%n+1])%mod;         }     }     memcpy(x,z,sizeof(z)); } void mont(long long t){     memset(ans,0,sizeof(ans));     ans[1]=1;    while(t>0){         if(t&1)multiply(ans,b);         t>>=1;         multiply(b,b);     } } int main(){while(scanf("%I64d%I64d%I64d%I64d",&n,&m,&d,&k)!=EOF){clear(a);clear(b);clear(ans);clear(z);long long i,j;for(i=1;i<=n;i++){scanf("%I64d",&a[i]);}for(i=1;i<=d+1;i++)b[i]=1;for(i=n;i>=n-d+1;i--)b[i]=1;mont(k);multiply(a,ans);for(i=1;i<=n;i++){printf("%I64d ",z[i]);}printf("\n");}}