Question

我遇到了另一个困难的Project Euler问题 Link to the problem

我的第一直觉是尝试一种简单的蛮力解决方案，这需要花费太多时间才能运行。

所以我想到了一个更好的解决方案，但我不知道如何编码。

我想：

生成所有必要的三元组。
汇总所有组合。
计算总和。

我做了第1步，我的结果看起来像这样：

Multiples of 17: [[0, 1, 7], [0, 3, 4], [0, 5, 1], [0, 6, 8], [0, 8, 5], [1,   0, 2], [1, 3, 6], [1, 5, 3], [1, 7, 0], [1, 8, 7], [2, 0, 4], [2, 3, 8], [2, 8, 9], [3, 0, 6], [3, 4, 0], [3, 5, 7], [3, 7, 4], [3, 9, 1], [4, 0, 8], [4, 2, 5], [4, 5, 9], [4, 7, 6], [4, 9, 3], [5, 1, 0], [5, 2, 7], [5, 6, 1], [5, 7, 8], [6, 1, 2], [6, 2, 9], [6, 8, 0], [6, 9, 7], [7, 1, 4], [7, 3, 1], [7, 4, 8], [7, 6, 5], [7, 8, 2], [8, 1, 6], [8, 5, 0], [8, 6, 7], [9, 0, 1], [9, 1, 8], [9, 3, 5], [9, 5, 2], [9, 8, 6]] etc...

现在，这对我来说是棘手的部分。我尝试将它们与嵌套循环放在一起，但这确实很混乱。如果您有任何建议，请告诉我：）

Answer 1

首先，蛮力解决方案几乎没有时间运行。

正如@MooingRawr建议的那样，如果你使用itertools.permutations，只有 ~0.9 x 9！ 0123456789 的排列不会以零。

from itertools import permutations

primes = [17, 13, 11, 7, 5, 3, 2]

total = 0

# Generate permutations of 10 distict digits -- 10 factorial
for i in permutations('0123456789'):

  # Discard those that begin with zero -- one-tenth of 10!
  if i[0] == '0':
    continue 

  # Convert into a string and assume that it is valid
  n = ''.join(list(i))
  valid = True

  # Check if the last three digits are divisible by 17, ...
  #    ... then shift left and check if those digits are divisible by 13, etc.
  for j in xrange(0, len(primes)):
    x = n[len(primes) - j:len(primes) - j + 3]
    if int(x) % primes[j]:
      valid = False
      break

  # Print and add
  if valid:
    print 'Valid number: %s' % n
    total += int(n)

print 'Total: %d' % total

如果您运行此解决方案here，它会在几秒钟内运行，这对于PE来说应该没问题。

但是，您建议的方法确实更有效。请注意，您已经硬编码了七个循环，我只是使用factors生成它们，其中factor[i]是 d_i d_i + 1 d_i + 2 的因子。

您担心生成所有组合，但这很简单，使用递归，其中每次迭代检查最后两位数字并找到有效的下一位数。

factors = [1, 2, 3, 5, 7, 11, 13, 17]
valid_len = len(factors)
valid_sequences  = []
total = 0

# Checks for a 3-digit number with 3 unique digits
def not_unique(digits):
  return (digits[0] == digits[1]) or (digits[1] == digits[2]) or (digits[0] == digits[2])


# For each of the prime numbers, generate all valid triples that have unique digits
for i in xrange(0 ,len(factors)):
  current_map = {}
  for j in xrange(factors[i], 1000, factors[i]):
    digits = str(j).zfill(3)

    # Prune those numbers that have non-unique digits
    if not_unique(digits):
      continue

    # current_map is of the form {'d1d2':[list of all possible valid d3s], ...}
    if digits[:2] not in current_map:
      current_map[digits[:2]] = [digits[2]]
    else:
      current_map[digits[:2]].append(digits[2])

  valid_sequences.append(current_map)


# Checks each triple starting with the 3 most significant digits
# Get the last two digits, and find all the valid values for the next one digit
# Perform recursively
def get_matches_starting_with(sequence, index):
  global total
  if index == valid_len:
    print 'Valid number: %s' % sequence
    total += int(sequence)
  else:
    pair = sequence[-2:]
    if pair in valid_sequences[index]:
      for digit in valid_sequences[index][pair]:
        if not digit in sequence:
          get_matches_starting_with(sequence + digit, index + 1)

all_matches = []
for pair in valid_sequences[0]:
  if pair[0] == '0':
      continue
  for digit in valid_sequences[0][pair]:
    triple = pair + digit
    all_matches.append(get_matches_starting_with(triple, 1))

print 'Total: %d' % total

您可能希望运行解决方案here，并可能在中间步骤打印值以查看正在进行的操作。

仍有很多机会修剪探索过的州的数量。你的方法将它从3265920降低到大约3000.

用Python编写euler 43项目

1 个答案: