http://hzwer.com/1817.html
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
|
|
<b>sample input #1</b>
5 3 1 1
1 2 2 1 2
<b>sample input #2</b>
5 3 1 10
1 2 2 1 2
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Sample Output
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<b>sample
output
#1</b>
2
2
2
2
1
<b>sample
output
#2</b>
2
0
0
2
2
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题解
首先来看一下Sample里的第一组数据。
1 2 2 1 2
经过一次变换之后就成了
5 5 5 5 4
它的原理就是
a0 a1 a2 a3 a4
->(a4+a0+a1) (a0+a1+a2) (a1+a2+a3) (a2+a3+a4) (a3+a4+a0)
如果用矩阵相乘来描述，那就可以表述为1xN和NxN的矩阵相乘，结果仍为1xN矩阵
b = 1 2 2 1 2
a =
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
b * a = 5 5 5 5 4
所以最终结果就是：a * (b^k)
复杂度是n^3logk，过不了
于是我们发现这个矩阵长得很奇葩，每一行都是上一行后移一位得到
所以我们每个矩阵可以变成一行，这样就可以少1个n
把原来的a[i][k]变成a[i-k]
然后就是快速幂。。。
#include<iostream>
#include<cstring>
#include<cstdio>
#define ll long long
using namespace std;
ll n,m,d,k;
ll num[500],a[500],b[500];
void prll(ll a[]){for(ll i=0;i<n;i++)cout<<a[i]<<' ';cout<<endl;}
void mul(ll a[],ll b[],ll ans[])
{
ll t[500];
memset(t,0,sizeof(t));
for(ll i=0;i<n;i++)
for(ll k=0;k<n;k++)
if(i-k>=0)t[i]=(t[i]+(a[k]*b[i-k])%m)%m;
else t[i]=(t[i]+(a[k]*b[i-k+n])%m)%m;
for(ll i=0;i<n;i++)ans[i]=t[i];
}
int main()
{
scanf("%d%d%d%d",&n,&m,&d,&k);
for(ll i=0;i<n;i++)scanf("%d",&num[i]);
for(ll i=0;i<=d;i++)a[i]=1;
for(ll i=n-1;i>=n-d;i--)a[i]=1;
b[0]=1;
while(k)
{
if(k&1)mul(b,a,b);
k>>=1;
mul(a,a,a);
}
mul(num,b,num);
prll(num);
return 0;
}
