Poj 3150 Cellular Automaton

Idea: rapid matrix Power Analysis: 1 The number of n given by the question is 0 ~ The distance between two numbers is min (| I-j |, n-| I-j | ). Given d and k, each number is converted into a new number after each transformation. This new number is equal to the sum of % m2, which is less than or equal to all numbers of d. This is similar to the previous two questions, hdu2276 and FZU1692, all of them belong to the problem of cyclic homogeneity. Let's first look at what each number becomes after a transformation. Because the distance is less than or equal to d, the first | I-j | = d, then j = I + d, the second case n-| I-j | = d, therefore, j = n-d + I. The first number equals to = num [1] + num [2] + .... + num [d + 1] + num [n-d + 1] +... + num [n] The second number equals to = num [2] + .... + num [d + 2] + num [n-d + 2] +... + num [n]... ........................................ ................................... 3 because the matrix here is cyclical and homogeneous, we only need to find the first line, and the rest can be introduced according to the previous line. In this way, the complexity of matrix multiplication is reduced to O (n ^ 2) code:

 /************************************************  * By: chenguolin                               *   * Date: 2013-08-31                             *  * Address: http://blog.csdn.net/chenguolinblog *  ************************************************/  #include<cstdio>  #include<cstring>  #include<iostream>  #include<algorithm>  using namespace std;    typedef long long int64;  const int N = 505;     int n , MOD , d , k;  int arr[N];    struct Matrix{      int64 mat[N][N];      Matrix operator*(const Matrix &m)const{          Matrix tmp;          for(int i = 1 ; i <= n ; i++){              tmp.mat[1][i] = 0;              for(int j = 1 ; j <= n ; j++)                  tmp.mat[1][i] += mat[1][j]*m.mat[j][i]%MOD;              tmp.mat[1][i] %= MOD;          }          for(int i = 2 ; i <= n ; i++){              tmp.mat[i][1] = tmp.mat[i-1][n];              for(int j = 2 ; j <= n ; j++)                  tmp.mat[i][j] = tmp.mat[i-1][(j-1+n)%n];          }          return tmp;       }  };    void init(Matrix &m){      memset(m.mat , 0 , sizeof(m.mat));      for(int i = 1 ; i <= d+1 ; i++)          m.mat[1][i] = 1;      for(int i = n-d+1 ; i <= n ; i++)          m.mat[1][i] = 1;      for(int i = 2 ; i <= n ; i++){          m.mat[i][1] = m.mat[i-1][n];          for(int j = 2 ; j <= n ; j++)              m.mat[i][j] = m.mat[i-1][(j-1+n)%n];      }  }    void Pow(Matrix &m){      Matrix ans;      memset(ans.mat , 0 , sizeof(ans.mat));      for(int i = 1 ; i <= n ; i++)          ans.mat[i][i] = 1;      while(k){          if(k&1)              ans = ans*m;          k >>= 1;          m = m*m;      }      for(int i = 1 ; i <= n ; i++){          int64 sum = 0;          for(int j = 1 ; j <= n ; j++)              sum += ans.mat[i][j]*arr[j]%MOD;          if(i > 1) printf(" ");          printf("%lld" , sum%MOD);      }      puts("");  }    int main(){      Matrix m;      while(scanf("%d" , &n) != EOF){          scanf("%d%d%d" , &MOD , &d , &k);          for(int i = 1 ; i <= n ; i++)              scanf("%d" , &arr[i]);          init(m);          Pow(m);      }          return 0;  }