Topic : http://poj.org/problem?id=3150
Cellular Automaton binary matrix multiplication +
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 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
The meaning of problems :
there are n number of a ring, it defines a method of operating it and the distance d is smaller than the sum of the number of modulo m. Each operation refresh all data. After asking k times will become what number?
Idea :
it is easy to think of a matrix structure 01, ksm solving, n too little optimization: find a regular matrix
:
1100001
1110000
0.111 million
.. . . .
Only need to keep the line of the first row, the multiplication is calculated as follows
for(int i=0;i<n;i++)
for(int j=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;
int n,m,d,k;
long long init[505],tmp[505];
void mul(long long a[],long long b[])
{
long long c[505];
for(int i=0;i<n;i++)
c[i]=0;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
c[i]+=a[j]*b[i>=j?(i-j):(n+i-j)];
for(int i=0;i<n;i++)
b[i]=c[i]%m;
}
int main()
{
int i,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");
}