"Poj 3150" Cellular automaton matrix

title : http://poj.org/problem?id=3150

Cellular Automaton matrix multiplication + dichotomy

Cellular Automaton

Time limit:12000ms Memory limit:65536k
Total submissions:3544 accepted:1428
Case Time Limit:2000ms

Description

A cellular automaton is a collection of cells in a grid of specified shape that evolves through a number of discrete time Steps according to a set of rules this 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 is 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 is considered to being integer numbers from 0 to M? 1.

One of the most fundamental properties of a cellular automaton are 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 O Rder m. We'll denote such kind of cellular automaton as N,m-automaton.

A distance between cells I and J in N,m-automaton are 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 is simultaneously replaced by new values. The new value of cell I after D-step was computed as a sum of values of cells belonging to the d-enviroment of the cell i m Odulo 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 ' 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

Test Instructions :
There is a number of n on a ring, which defines an operation that is less than the number of D plus and re-modulo m. Refreshes all numbers for each operation. What number will it be after asking K times?

Ideas :
It's easy to think of constructing a 01 matrix, KSM solving, n too big a bit of optimization: Discovering that the matrix is regular
：
1100001
1110000
0111000
.。。。。
Just need to save the first row, the multiplication is calculated as follows

fori=0;i<n;i++)      forj=0;j<n;j++)      c[i]+=a[j]*b[i>=j?(i-j):(n+i-j)];

Code :

#include <iostream>#include <stdio.h>using namespace STD;intN,m,d,k;Long Longinit[505],tmp[505];voidMulLong LongA[],Long LongB[]) {Long Longc[505]; for(intI=0; i<n;i++) c[i]=0; for(intI=0; i<n;i++) for(intj=0; j<n;j++) C[i]+=a[j]*b[i>=j? ( I-J):(n+i-j)]; for(intI=0; i<n;i++) b[i]=c[i]%m; }intMain () {intI,j;scanf("%d%d%d%d", &n,&m,&d,&k); for(i=0; i<n;++i)scanf("%lld", &init[i]); for(tmp[0]=i=1; i<=d;++i) tmp[i]=tmp[n-i]=1; while(k) {if(k&1) Mul (Tmp,init); k>>=1;    Mul (TMP,TMP); } for(i=0; i<n;i++)printf("%lld", Init[i]);printf("\ n");}

"Poj 3150" Cellular automaton matrix