FZU 1692 Key problem

Idea: constructor matrix + rapid Power Analysis: 1 The question means that there are n people in a circle, and each person initially has an ai apple. Now I am playing m games, after each game, I personally can add R * A (I + N-1) % n + L * A (I + 1) % n Apples (wrong question ), q: After the m-round game, the number of apples for each person 2. Based on the question, we can list the number of apples for each person after the round. a0 = a0 + R * an-1 + L * a1 a1 = a1 + R * a0 + L * a2 ............................. an-1 = an-1 + R * an-2 + L * a03 based on the second idea, we can construct the following matrix 1 L 0 ...... R a0 a0 'r 1 L ......... * a1 '....................... = ................. R 1 L an-2 an-2 'l ........... R 1 an-1 an-1' 4 then we can use the matrix power to quickly obtain the final answer based on 3, but the n of the question is 100 at the maximum, and m is 10 ^ 9 at the maximum, therefore, the time complexity of each case is O (Logm * n ^ 3). When n is 100 at most, we can see 5 of TLE in the initial matrix, A matrix is a cyclical homogeneous structure, that is, each row of the matrix can be pushed from the previous row. Therefore, we only need to use the time of O (n ^ 2) to find the first row, then we are using recursion to find the remaining n-1 rows, so the total time complexity is O (Logm * n ^ 2) code:

/************************************************  * By: chenguolin                               *   * Date: 2013-08-30                             *  * Address: http://blog.csdn.net/chenguolinblog *  ************************************************/  #include<cstdio>  #include<cstring>  #include<iostream>  #include<algorithm>  using namespace std;    typedef __int64 int64;  const int N = 110;    int arr[N];  int n , m , L , R , MOD;    struct Matrix{      int64 mat[N][N];      Matrix operator*(const Matrix& ma)const{          Matrix tmp;          for(int i = 0 ; i < n ; i++){              tmp.mat[0][i] = 0;               for(int j = 0 ; j < n ; j++)                  tmp.mat[0][i] += mat[0][j]*ma.mat[j][i]%MOD;              tmp.mat[0][i] %= MOD;          }          for(int i = 1 ; i < n ; i++)              for(int j = 0 ; j < n ; j++)                  tmp.mat[i][j] = tmp.mat[i-1][(j-1+n)%n];           return tmp;      }  };    void init(Matrix &ma){      memset(ma.mat , 0 , sizeof(ma.mat));      ma.mat[0][1] = L ; ma.mat[0][n-1] = R;      ma.mat[n-1][0] = L ; ma.mat[n-1][n-2] = R;      ma.mat[0][0] = ma.mat[n-1][n-1] = 1;      for(int i = 1 ; i < n-1 ; i++){          ma.mat[i][i-1] = R;          ma.mat[i][i+1] = L;          ma.mat[i][i] = 1;      }  }    void Pow(Matrix &ma){      Matrix ans;      memset(ans.mat , 0 , sizeof(ans.mat));      for(int i = 0 ; i < n ; i++)          ans.mat[i][i] = 1;      while(m){          if(m&1)              ans = ans*ma;          m >>= 1;          ma = ma*ma;      }      for(int i = 0 ; i < n ; i++){          int64 sum = 0;          for(int j = 0 ; j < n ; j++)              sum += ans.mat[i][j]*arr[j]%MOD;          if(i) printf(" ");          printf("%I64d" , sum%MOD);      }       puts("");  }    int main(){      int cas;      Matrix ma;      scanf("%d" , &cas);      while(cas--){          scanf("%d%d%d" , &n , &m , &L);           scanf("%d%d" , &R , &MOD);           for(int i = 0 ; i < n ; i++)              scanf("%d" , &arr[i]);          init(ma);          Pow(ma);      }      return 0;  }