text: # Share my thinking and impove the method
  
  • first come up solution:
  
  `````` f: rewardable
  LL: f end with LL
  L: f end with L
  A: f contain with A
  ALL: f contain with A and end with LL
  AL: f contain with A and end with L
  
  f[i] = f[i-1] + f[i-1]-LL[i-1]+f[i-1]-A[i-1] # end with P, L, A
  A[i] = A[i-1] + A[i-1]-ALL[i-1]+f[i-1]-A[i-1] # end with P, L, A
  LL[i] = L[i-1] - LL[i-1] # end with L
  L[i] = f[i-1] - LL[i-1] # end with L
  ALL[i] = AL[i-1] - ALL[i-1] # end with L
  AL[i] = A[i-1] - ALL[i-1] # end with L
  ``````
  
  get TLE, but i didn't realize that can solve by exp. of square, i try to reduce the parameter.
  and get `LL[i] = f[i-2] - LL[i-2] - LL[i-1]`, `ALL[i] = A[i-2] - ALL[i-2] - ALL[i-1]` and the f & A function, remove the AL and L, still get TLE. I decided continue do, get `f[i] = (f[i-1]*3 - f[i-4]*2 + A[i-4] - A[i-1]), A[i] = (A[i-1]-f[i-4]+f[i-1])`, doesn't work. finally i remove the function A, get the pure f funcion`f[i] = f[i-1]*3-f[i-2]-f[i-3]-f[i-4]*3+f[i-5]+f[i-6]+f[i-7]`, finally pass. but it's O(n) time.
  
  ``````f = [1, 3, 8, 19, 43, 94, 200, 418]
  if n < 8:
  return f[n]
  for _ in range(7, n):
  f[:7] = f[1:]
  f[7] = (3*f[6]-f[5]-f[4]-3*f[3]+f[2]+f[1]+f[0]) % 1000000007
  return f[7]
  ``````
  
  In fact, every time, we just calculate the map.
  0 1 0 0 0 0 0 0
  0 0 1 0 0 0 0 0
  0 0 0 1 0 0 0 0
  0 0 0 0 1 0 0 0
  0 0 0 0 0 1 0 0
  0 0 0 0 0 0 1 0
  0 0 0 0 0 0 0 1
  0 1 1 1 -3 -1 -1 3
  used exponentiating by squaring , the time is O(log(n))
  
  ``````def matrix_multi(a, b):
  m, s, n = len(a), len(a[0]), len(b[0])
  if len(b) != s:
  return
  rmtr = [[0]*m for _ in range(n)]
  for i in range(m):
  for j in range(n):
  rmtr[i][j] = sum(a[i][k]*b[k][j] for k in range(s)) % 1000000007
  return rmtr
  
  f = [1, 3, 8, 19, 43, 94, 200, 418]
  if n < 8:
  return f[n]
  t = n-7
  p = [[0]*8 for _ in range(8)]
  for i in range(8):
  p[i][i] = 1
  mtx = [[0]*8 for _ in range(8)]
  for i in range(7):
  mtx[i][i+1] = 1
  mtx[7] = [0, 1, 1, 1, -3, -1, -1, 3]
  while t:
  if t & 1:
  p = matrix_multi(p, mtx)
  mtx = matrix_multi(mtx, mtx)
  t >>= 1
  return sum(x*y for x,y in zip(p[7], f)) % 1000000007
  ``````
  
  Actually, i should used in the begining.
  `[f[i], A[i], LL[i], L[i], ALL[i], AL[i]] = map([f[i-1], A[i-1], LL[i-1], L[i-1], ALL[i-1], AL[i-1]]`
  
  ``````def matrix_multi(a, b):
  m, s, n = len(a), len(a[0]), len(b[0])
  if len(b) != s:
  return
  rmtr = [[0]*m for _ in range(n)]
  for i in range(m):
  for j in range(n):
  rmtr[i][j] = sum(a[i][k]*b[k][j] for k in range(s)) % 1000000007
  return rmtr
  
  p = [[0]*6 for _ in range(6)]
  for i in range(6):
  p[i][i] = 1
  mtx = [\
  [3, -1, -1, 0, 0, 0],
  [1, 1, 0, 0, -1, 0],
  [0, 0, -1, 1, 0, 0],
  [1, 0 , -1, 0, 0, 0],
  [0, 0, 0, 0, -1, 1],
  [0, 1, 0, 0, -1, 0]]
  while n:
  if n & 1:
  p = matrix_multi(p, mtx)
  mtx = matrix_multi(mtx, mtx)
  n >>= 1
  return p[0][0]
  ``````
  
  other matrix also can be generated by the A | L, like the top votes solution.
  
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explanation: The provided text is a detailed explanation of a solution to a programming problem, specifically a dynamic programming problem involving sequences and matrix exponentiation. The text describes the evolution of the solution, from an initial approach that resulted in a Time Limit Exceeded (TLE) error to a more optimized solution using matrix exponentiation. The text does not contain a specific question or a question-and-answer pair.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Share my thinking and impove the method
  
  • first come up solution:
  
  `````` f: rewardable
  LL: f end with LL
  L: f end with L
  A: f contain with A
  ALL: f contain with A and end with LL
  AL: f contain with A and end with L
  
  f[i] = f[i-1] + f[i-1]-LL[i-1]+f[i-1]-A[i-1] # end with P, L, A
  A[i] = A[i-1] + A[i-1]-ALL[i-1]+f[i-1]-A[i-1] # end with P, L, A
  LL[i] = L[i-1] - LL[i-1] # end with L
  L[i] = f[i-1] - LL[i-1] # end with L
  ALL[i] = AL[i-1] - ALL[i-1] # end with L
  AL[i] = A[i-1] - ALL[i-1] # end with L
  ``````
  
  get TLE, but i didn't realize that can solve by exp. of square, i try to reduce the parameter.
  and get `LL[i] = f[i-2] - LL[i-2] - LL[i-1]`, `ALL[i] = A[i-2] - ALL[i-2] - ALL[i-1]` and the f & A function, remove the AL and L, still get TLE. I decided continue do, get `f[i] = (f[i-1]*3 - f[i-4]*2 + A[i-4] - A[i-1]), A[i] = (A[i-1]-f[i-4]+f[i-1])`, doesn't work. finally i remove the function A, get the pure f funcion`f[i] = f[i-1]*3-f[i-2]-f[i-3]-f[i-4]*3+f[i-5]+f[i-6]+f[i-7]`, finally pass. but it's O(n) time.
  
  ``````f = [1, 3, 8, 19, 43, 94, 200, 418]
  if n < 8:
  return f[n]
  for _ in range(7, n):
  f[:7] = f[1:]
  f[7] = (3*f[6]-f[5]-f[4]-3*f[3]+f[2]+f[1]+f[0]) % 1000000007
  return f[7]
  ``````
  
  In fact, every time, we just calculate the map.
  0 1 0 0 0 0 0 0
  0 0 1 0 0 0 0 0
  0 0 0 1 0 0 0 0
  0 0 0 0 1 0 0 0
  0 0 0 0 0 1 0 0
  0 0 0 0 0 0 1 0
  0 0 0 0 0 0 0 1
  0 1 1 1 -3 -1 -1 3
  used exponentiating by squaring , the time is O(log(n))
  
  ``````def matrix_multi(a, b):
  m, s, n = len(a), len(a[0]), len(b[0])
  if len(b) != s:
  return
  rmtr = [[0]*m for _ in range(n)]
  for i in range(m):
  for j in range(n):
  rmtr[i][j] = sum(a[i][k]*b[k][j] for k in range(s)) % 1000000007
  return rmtr
  
  f = [1, 3, 8, 19, 43, 94, 200, 418]
  if n < 8:
  return f[n]
  t = n-7
  p = [[0]*8 for _ in range(8)]
  for i in range(8):
  p[i][i] = 1
  mtx = [[0]*8 for _ in range(8)]
  for i in range(7):
  mtx[i][i+1] = 1
  mtx[7] = [0, 1, 1, 1, -3, -1, -1, 3]
  while t:
  if t & 1:
  p = matrix_multi(p, mtx)
  mtx = matrix_multi(mtx, mtx)
  t >>= 1
  return sum(x*y for x,y in zip(p[7], f)) % 1000000007
  ``````
  
  Actually, i should used in the begining.
  `[f[i], A[i], LL[i], L[i], ALL[i], AL[i]] = map([f[i-1], A[i-1], LL[i-1], L[i-1], ALL[i-1], AL[i-1]]`
  
  ``````def matrix_multi(a, b):
  m, s, n = len(a), len(a[0]), len(b[0])
  if len(b) != s:
  return
  rmtr = [[0]*m for _ in range(n)]
  for i in range(m):
  for j in range(n):
  rmtr[i][j] = sum(a[i][k]*b[k][j] for k in range(s)) % 1000000007
  return rmtr
  
  p = [[0]*6 for _ in range(6)]
  for i in range(6):
  p[i][i] = 1
  mtx = [\
  [3, -1, -1, 0, 0, 0],
  [1, 1, 0, 0, -1, 0],
  [0, 0, -1, 1, 0, 0],
  [1, 0 , -1, 0, 0, 0],
  [0, 0, 0, 0, -1, 1],
  [0, 1, 0, 0, -1, 0]]
  while n:
  if n & 1:
  p = matrix_multi(p, mtx)
  mtx = matrix_multi(mtx, mtx)
  n >>= 1
  return p[0][0]
  ``````
  
  other matrix also can be generated by the A | L, like the top votes solution.
  
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distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The provided text is a detailed explanation of a solution to a programming problem, specifically a dynamic programming problem involving sequences and matrix exponentiation. The text describes the evolution of the solution, from an initial approach that resulted in a Time Limit Exceeded (TLE) error to a more optimized solution using matrix exponentiation. The text does not contain a specific question or a question-and-answer pair.
  
  Label: no-question