Question

似乎无法弄清楚我的代码有什么问题（Project Euler问题8）。我想找到下面10,000位数字中13个相邻数字的最大乘积，我的答案是错误的。

my_list = list('7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450')

for i in range(1000):
    my_list[i] = int(my_list[i])

previous_product = 1
for x in range(13):
  previous_product *= my_list[x]
current_product = previous_product*my_list[13]/my_list[0]

for i in range(1, 987):
  if current_product > previous_product:
    maximum_product = current_product
  previous_product = current_product
  if my_list[i]==0:
    current_product = 1
    for x in range(13):
      current_product *= my_list[i+x+1]
  else:
    current_product = previous_product*my_list[i+x+1]/my_list[i]

print(maximum_product)

编辑：解决了！ maximum_product的定义错误...它取代了最新“当前产品”的价值，该价值恰好大于以前的产品，不一定是最大的产品。

正确，尽管不是超级高效的代码：

   my_list = list('7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450')

for i in range(1000):
    my_list[i] = int(my_list[i])

previous_product = 1
for x in range(13):
  previous_product *= my_list[x]
current_product = previous_product*my_list[13]/my_list[0]

large_products = []

for i in range(1, 987):
  if current_product > previous_product:
    large_products.append(current_product)
  previous_product = current_product
  if my_list[i]==0:
    current_product = 1
    for x in range(13):
      current_product *= my_list[i+x+1]
  else:
    current_product = previous_product*my_list[i+x+1]/my_list[i]

print(max(large_products))

Answer 1

我还没有检查为什么你的方法不起作用，但你可以通过以下方式轻松解决这个问题：

import operator
from functools import reduce

my_int_list = [int(char) for char in my_list]

max(map(lambda *x: reduce(operator.mul, x), 
        my_int_list[0:], 
        my_int_list[1:], 
        my_int_list[2:], 
        my_int_list[3:], 
        my_int_list[4:], 
        my_int_list[5:], 
        my_int_list[6:], 
        my_int_list[7:], 
        my_int_list[8:], 
        my_int_list[9:], 
        my_int_list[10:], 
        my_int_list[11:], 
        my_int_list[12:]))

如果这会占用太多内存，您也可以使用itertools.islice代替使用[idx:]进行直接切片。

Answer 2

以下是您的滑动窗口构思的实现，已修改，以便它仅适用于不包含零的字符串：

def max_product(s,k):
    s = [int(d) for d in s]
    p = 1
    for d in s[:k]:
        p *= d
    m = p
    for i,d in enumerate(s[k:]):
        p *= d
        p //= s[i]
        if p > m: m = p
    return p

在上面的s中，长度至少为k（问题中为k = 13）的非零数字字符串。它返回k个连续数字的最大乘积。微妙的方式是枚举作品。当您在enumerate上使用s[k:]时，i的索引从0开始 - 这正是您想要在第一次通过该循环时删除的因素。

将此应用于

data = '7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450'

首先拆分为不包含零的块，且长度至少为13位：

chunks = [s for s in data.split('0') if len(s) >= 13]

有24个这样的块。要获得总体最大值，只需占用每个块的最大值：

print(max(max_product(s,13) for s in chunks))

确实打印23514624000

Answer 3

您当前的代码无效，因为它似乎是基于错误的计划 - 或者至少是我不完全理解的计划。这是一种不同的方式考虑算法。

想象一下，我们有这个号码：7316717。我们正在寻找最大值 3个相邻数字的乘积。我们可以解决这个问题如下。我对每个步骤的结果进行硬编码，但您应该能够编写Python 用于计算每个部分的代码。

查找所有相邻的3位数序列：
```
seqs = [731, 316, 167, 671, 717]
```
计算他们的产品：
```
products = [21, 18, 42, 42, 49]
```
找到他们的最大值。
```
answer = max(products)
```

Answer 4

我还没有检查过你的代码，以及它以何种方式工作。

但是解决问题的一个简单方法是，从输入数据生成所有可能的切片，其中包含13个相邻数字，然后找到其元素的乘积然后找到它们的最大乘积。

这是一种方法：

data = '7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450'

def product(a):
    prod = 1
    for k in a:
        prod *= int(k)
    return prod

def max_adjacent_number(data, suit=13):
    return max(product(data[k:k+suit]) for k in range(len(data)) if '0' not in data[k:k+suit])


from time import time
start = time()
val = max_adjacent_number(data)
elapsed = time() - start
print("Solution: {0} \telapsed: {1:.5f}ms".format(val, elapsed*1000))

输出：

# Best time
Solution: 23514624000   elapsed: 1.31226ms

Answer 5

您可以使用以下列表理解来提高代码效率：

import time
n = str(x) #x is the long number
a = time.time()
result = max(reduce(lambda x, y: x * y, map(int, n[i:i+13])) for i in xrange(len(n)-12) if '0' not in n[i:i+13])
b = time.time()
print "Maximum adjacent numbers product: %d" % result
print "Time taken:", (b-a)*1000, "ms"

输出：

Maximum adjacent numbers product: 23514624000
Time taken: 4.34899330139 ms

查找10,000位数字

5 个答案: