POJ 3150/ UVA 1386 Cellular Automaton（矩阵乘法&循环矩阵）

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

Source

Northeastern Europe 2006

题意：给你一个循环的数组，经过一次变换，每一个数变成原来数组里自己和距离自己不超过d的数字的和模m。

求k次变换后数组的值。

对于第一个样例，可以很轻松的构造出矩阵：

A： 

1 2 2 1 2

B：

1 1 0 0 1

1 1 1 0 0 

0 1 1 1 0

0 0 1 1 1 

1 0 0 1 1

然后答案就是A*（B^k），然而，n有500那么大，复杂度就是O(n^3*log(k))，所以会超时。

但是我们观察发现B是一个循环矩阵，循环矩阵有一个性质 如果A是循环矩阵，那么A^2, A^3.....都是循环矩阵。

那么我们做矩阵乘法的时候就只用算出矩阵的第一行，下面每一行都可以由上一行推出，那么这样复杂度都变成了O(n^2*log(k)).

还有就是500*500在结构体里开不下，无奈只能用普通数组。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define LL long longusing namespace std;int n,m,d,k;LL A[500][500],B[500][500],ans[500][500],t[500][500];void mul(LL a[500][500],LL b[500][500],LL ans[500][500]){    int i,j;    memset(t,0,sizeof(t));    for(i=0;i<n;i++)        for(j=0;j<n;j++)        {            t[0][i] += a[0][j]*b[j][i];            t[0][i] %= m;        }    for(i=1;i<n;i++)        for(j=0;j<n;j++)            t[i][j] = t[i-1][(j-1+n)%n];    for(i=0;i<n;i++)        for(j=0;j<n;j++)            ans[i][j] = t[i][j];}void pow(LL a[500][500],int k){    int i,j;    for(i=0;i<n;i++)        for(j=0;j<n;j++)            ans[i][j] = (i==j);        while(k)        {            if(k&1)                mul(ans,B,ans);            mul(B,B,B);            k /= 2;        }    for(i=0;i<n;i++)        for(j=0;j<n;j++)            a[i][j] = ans[i][j];}int main(void){    int i,j;    while(scanf("%d%d%d%d",&n,&m,&d,&k)==4)    {        memset(A,0,sizeof(A));        memset(B,0,sizeof(B));        for(i=0;i<n;i++)            scanf("%lld",&A[0][i]);        for(j=0;j<n;j++)        {            for(i=0;i<=d;i++)            {                B[(j+i)%n][j] = 1;                B[(j-i+n)%n][j] = 1;            }        }        pow(B,k);        mul(A,B,ans);        for(i=0;i<n-1;i++)            printf("%d ",ans[0][i]);        printf("%d\n",ans[0][i]);    }    return 0;}