Python 在10000位数字中查找13个相邻数字的最大乘积
似乎无法找出我的代码出了什么问题(Project Euler问题8)。我想在下面的10000位数字中找到13个相邻数字的最大乘积,但我得到了错误的答案Python 在10000位数字中查找13个相邻数字的最大乘积,python,Python,似乎无法找出我的代码出了什么问题(Project Euler问题8)。我想在下面的10000位数字中找到13个相邻数字的最大乘积,但我得到了错误的答案 my_list = list('73167176531330624919225119674426574742355349194934969835203127745063262395783180169848018694788518438586156078911294949545950173795833195285320880551112540698
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!最大_乘积定义错误。。。它承担了最近的“当前产品”的价值,该价值恰好大于先前的产品,而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))
我没有检查您的方法不起作用的原因,但您可以通过以下方法轻松解决此问题:
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:]))
如果这占用了太多内存,您也可以使用
itertools.islice
而不是使用[idx:
直接切片,因为您当前的代码似乎基于错误的计划,因此无法正常工作
--或者至少是一个我不完全理解的计划。这是另一种方式
考虑一下算法
假设我们有这个号码:7316717
。我们正在寻找最大值
三个相邻数字的乘积。我们可以如下解决这个问题。我是
硬编码每个步骤的结果,但您应该能够编写Python
计算每个部分的代码
- 查找所有相邻的3位序列:
seqs = [731, 316, 167, 671, 717]
- 计算他们的产品:
products = [21, 18, 42, 42, 49]
- 找到它们的最大值
answer = max(products)
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
Maximum adjacent numbers product: 23514624000
Time taken: 4.34899330139 ms
下面是滑动窗口思想的一个实现,经过修改,它只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,是一个长度至少为k
的非零数字字符串(在您的问题中为k=13
)。它返回k
连续数字的最大乘积。精妙之处在于enumerate的工作方式。当您在s[k://code>上使用enumerate
时,索引i
从0
开始,这正是您在第一次通过该循环时要删除的因子
适用于
data = '7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450'
首先分为不包含零且长度至少为13位的块:
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,只需获取每个块的最大值中的最大值:
print(max(max_product(s,13) for s in chunks))
它确实可以打印23514624000
您可以使用如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:
# Best time
Solution: 23514624000 elapsed: 1.31226ms
Maximum adjacent numbers product: 23514624000
Time taken: 4.34899330139 ms
这个问题可能属于。为什么不在0
上拆分,分别处理每一部分,每个部分的逻辑都要简单得多?另外,如果这是Python 3,您需要了解/
与/
之间的区别。您的代码以什么方式不工作?@Teodor:CR代表工作代码:请阅读。@martineau它给出了错误的答案:313528320.0而不是23514624000OP的方法是使用13位数的滑动窗口,通过乘以进入窗口的下一个数字,再除以离开窗口的数字来更新window@user3080953我懂了。好吧,这可能行得通,但可能很棘手,因为需要一些额外的逻辑来处理零:当一个人进入滑动窗口时,你的运行产品变成零(这很容易),但当零离开窗口时,你需要重新计算运行产品。窗口中可能有多个零。听起来我很容易犯错误。是的,伙计,很棒的解决方案,事实上你可以用islice和另一个mapmax(map(lambda*x:reduce(operator.mul,x),*map(lambda x:islice(my_int_list,x,None),xrange(13))
或者这个:max(reduce(mul,xs[i:i+13],1)表示范围内的i(0,len(xs)-13+1))