2012年亚洲区域赛东京站

C. One-Dimensional Cellular Automaton

Time Limit: 3000ms

Memory Limit: 131072KB
64-bit integer IO format: %lld
Java class name: Main

Submit Status PID:
33681

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There is a one-dimensional cellular automaton consisting of N cells. Cells are numbered from 0 to N -
1.
Each cell has a state represented as a non-negative integer less than M. The states of cells evolve through
discrete time steps. We denote the state of the i-th cell at time t as S(i, t).
The state at time t + 1 is defined by the equation

S(i, t + 1) = A x S(i - 1, t) + B x S(i, t) + C x S(i + 1, t) mod M,

(1)

where A, B and C are
non-negative integer constants. For i < 0 or N

i,
we define S(i, t) = 0.
Given an automaton definition and initial states of cells, your mission is to write a program that computes the states of the cells at a specified time T.

Input

The input is a sequence of datasets. Each dataset is formatted as follows.

N M A B C T

S(0, 0) S(1, 0) ... S(N - 1, 0)

The first line of a dataset consists of six integers, namely N, M, A, B, C and T. N is
the number of cells. M is the modulus in the equation (1). A, B and C are
coefficients in the equation (1). Finally, T is the time for which you should compute the states.
You may assume that 0 < N

50, 0
< M

1000, 0

A, B, C < M and 0

T

109.
The second line consists of N integers, each of which is non-negative and less than M.
They represent the states of the cells at time zero.
A line containing six zeros indicates the end of the input.

Output

For each dataset, output a line that contains the states of the cells at time T. The format of the output
is as follows.

S(0, T) S(1, T) ... S(N - 1, T)

Each state must be represented as an integer and the integers must be separated by a space.

Sample Input

5 4 1 3 2 0
0 1 2 0 1
5 7 1 3 2 1
0 1 2 0 1
5 13 1 3 2 11
0 1 2 0 1
5 5 2 0 1 100
0 1 2 0 1
6 6 0 2 3 1000
0 1 2 0 1 4
20 1000 0 2 3 1000000000
0 1 2 0 1 0 1 2 0 1 0 1 2 0 1 0 1 2 0 1
30 2 1 0 1 1000000000
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
30 2 1 1 1 1000000000
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
30 5 2 3 1 1000000000
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0

Sample Output

0 1 2 0 1
2 0 0 4 3
2 12 10 9 11
3 0 4 2 1
0 4 2 0 4 4
0 376 752 0 376 0 376 752 0 376 0 376 752 0 376 0 376 752 0 376
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0
1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 3 2 2 2 3 3 1 4 3 1 2 3 0 4 3 3 0 4 2 2 2 2 1 1 2 1 3 0

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
using namespace std;
const int maxn=55;
int map[maxn][maxn][maxn];
int num[maxn];//存储N个数据
int flag_1[maxn][maxn];
int flag_2[maxn];
int N;//含义如题
int M;
int A;
int B;
int C;
int T;
int main()
{
while(~scanf("%d%d%d%d%d%d",&N,&M,&A,&B,&C,&T))
{
if(N==0&&M==0&&A==0&&B==0&&C==0&&T==0)
break;
for(int i=1; i<=N; i++) //接收数据
scanf("%d",&num[i]);
if(T==0)//特判
{
printf("%d",num[1]);
for(int i=2; i<=N; i++)
printf(" %d",num[i]);
printf("\n");
continue;
}
for(int i=1; i<=N; i++)
{
map[0][i-1][i]=A;//存储系数
map[0][i][i]=B;
map[0][i+1][i]=C;
}
int len=0;//长度
while(T>1)
{
flag_2[++len]=T%2;
T=T/2;
}
for(int i=1; i<=len/2; i++)
{
int temp=flag_2[i];
flag_2[i]=flag_2[len-i+1];
flag_2[len-i+1]=temp;
}
for(int l=1; l<=len; l++)
{
for(int i=1; i<=N; i++)
for(int j=1; j<=N; j++)
map[l][i][j]=0;
for(int k=1; k<=N; k++)
for(int i=1; i<=N; i++)
for(int j=1; j<=N; j++)
map[l][i][j]=(map[l][i][j]+map[l-1][i][k]*map[l-1][k][j])%M;
if(flag_2[l]==1)
{
for(int i=1; i<=N; i++)
for(int j=1; j<=N; j++)
{
flag_1[i][j]=map[l][i][j];
map[l][i][j]=0;
}
for(int k=1; k<=N; k++)
for(int i=1; i<=N; i++)
for(int j=1; j<=N; j++)
map[l][i][j]=(map[l][i][j]+flag_1[i][k]*map[0][k][j])%M;
}
}
memset(flag_1,0,sizeof(flag_1));
for(int k=1; k<=N; k++)
for(int i=1; i<=N; i++)
for(int j=1; j<=N; j++)
flag_1[i][j]=(flag_1[i][j]+num[k]*map[len][k][j])%M;
printf("%d",flag_1[1][1]);
for(int i=2; i<=N; i++)
printf(" %d",flag_1[1][i]);
printf("\n");
}
return 0;
}

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <math.h>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#include <algorithm>
#include <set>
using namespace std;
int mod;
struct lt
{
int m[55][55];
int l;
lt operator *(lt b)
{
lt c(l);
for(int i=0; i<l; ++i)
for(int j=0; j<l; ++j)
{
if(m[j][i])
for(int k=0; k<l; ++k)
{
c.m[j][k]+=m[j][i]*b.m[i][k]%mod;
c.m[j][k]%=mod;
}
}
return c;
}
lt(int len)
{
l=len;
memset(m,0,sizeof(m));
}
};
lt ksm(lt a,int b)
{
lt c(a.l);
for(int i=0; i<c.l; ++i)
c.m[i][i]=1;
while(b)
{
if(b&1)
c=c*a;
a=a*a;
b>>=1;
}
return c;
}
int main()
{
int n,a,b,c,t;
while(~scanf("%d%d%d%d%d%d",&n,&mod,&a,&b,&c,&t))
{
if(!n&&!mod&&!a&&!b&&!c&&!t)
break;
lt m0(n),temp(n);
for(int i=0; i<n; ++i)
scanf("%d",&m0.m[0][i]);
for(int i=0; i<n; ++i)
{
if(i)
temp.m[i-1][i]=a;
temp.m[i][i]=b;
if(i+1<n)
temp.m[i+1][i]=c;
}
temp=ksm(temp,t);
lt mt(n);
mt=m0*temp;
for(int i=0; i<n; ++i)
if(i)
printf(" %d",mt.m[0][i]);
else
printf("%d",mt.m[0][i]);
printf("\n");
}
return 0;
}