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 < n2 , 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
Source Northeastern Europe 2006
容易推出转移矩阵,但是规模太大,还不行 观察发现这个矩阵的每一行都是循环的,下一行的就是上一行向右循环移动一位,所以我们可以把矩阵乘法降到n^2,只用一维数组来保存,这样再结合矩阵快速幂就可以ac了
/*************************************************************************> File Name: POJ3150.cpp> Author: ALex> Mail: zchao1995@gmail.com> Created Time: 2015年03月17日 星期二 20时42分20秒 ************************************************************************/;const double pi = acos(-1.0);const int inf = 0x3f3f3f3f;const double eps = 1e-15;LL;typedef pair <int, int> PLL;LL mat[505][505];LL base[505];LL arr[505];LL tmp[505];LL init[505];int n, m, d, k;void mul (LL a[]){for (int j = 1; j <= n; ++j){tmp[j] = 0;for (int i = 1; i <= n; ++i){tmp[j] += mat[i][j] * a[i];tmp[j] %= m;}}for (int i = 1; i <= n; ++i){a[i] = tmp[i];}}void fastpow(){while (k){if (k & 1){mul (arr);}k >>= 1;mul (base);for (int i = 1; i <= n; ++i){for (int j = 1; j <= n; ++j){int p = (j + i – 1) % n;if (!p){p = n;}mat[i][p] = base[j];}}}}int main (){while (~scanf(“%d%d%d%d”, &n, &m, &d, &k)){for (int i = 1; i <= n; ++i){scanf(“%lld”, &init[i]);}for (int j = 1; j <= n; ++j){for (int i = 1; i <= n; ++i){int dis = min (abs(i – j), n – abs(i – j));if (dis <= d){mat[i][j] = 1;}}}memset (arr, 0, sizeof(arr));arr[1] = 1; //单位矩阵for (int i = 1; i <= n; ++i){base[i] = mat[1][i];}fastpow();for (int i = 1; i <= n; ++i){for (int j = 1; j <= n; ++j){int p = (j + i – 1) % n;if (!p){p = n;}mat[i][p] = arr[j];}}mul(init);for (int i = 1; i <= n; ++i){printf(“%lld”, init[i]);if (i < n){printf(” “);}}printf(“\n”);}return 0;}
，我不去想是否能够成功，既然选择了远方，便只顾风雨兼程！
