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将数组中的元素相乘(Python)

  •  2
  • 78282219  · 技术社区  · 7 年前

    我想问一个问题,我需要找到一个长数字中包含的两位数的最大乘积:

    Number = 73167176531330624919225119674426574742355349194934
    
    lst = [int(i) for i in str(Number)]
    print(lst)
    
    new_lst = []
    
    for i in lst:
        new_lst.append(i*(i+1))
        print(new_lst)
    

    然而,我在上面尝试的结果是56,我如何乘以7*3,然后乘以3*1,依此类推。。。

    为H编辑

    Number = 7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450
    
    lst = [int(i) for i in str(Number)]
    print(lst)
    
    new_lst = []
    
    for i, value in enumerate(lst):
        new_lst.append(lst[i-13]*(lst[i-12])*(lst[i-11])*(lst[i-10])*(lst[i-9])*(lst[i-8])*(lst[i-7])*(lst[i-6])*(lst[i-5])*(lst[i-4])*(lst[i-3])*(lst[i-2])*(lst[i-1]))
    print(max(new_lst))
    
    7 回复  |  直到 7 年前
        1
  •  2
  •   blhsing    7 年前

    你可以迭代一个 zip 列表及其开头相差1的部分:

    max(a * b for a, b in zip(lst, lst[1:]))
    

    这将返回: 54 9 6 (数字中)

        2
  •  3
  •   jpp    7 年前

    您当前正在将数字i乘以i+1。你可以用 range 做你想做的事:

    for i in range(len(lst)-1):
            new_lst.append(lst[i]*(lst[i+1]))
            print(new_lst)
    
        3
  •  1
  •   timgeb    7 年前

    itertools 有一种方法叫 pairwise

    >>> from itertools import tee
    >>> def pairwise(iterable):
    ...     "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    ...     a, b = tee(iterable)
    ...     next(b, None)
    ...     return zip(a, b)
    ... 
    >>> [a*b for a,b in pairwise(map(int, '1234'))]
    [2, 6, 12]
    
        4
  •  1
  •   Coder.Yyy    7 年前

    号码=73167176531330624919225119674265742355349194934

    lst=[int(i)表示str(Number)中的i]

    新建\u lst=[]

    新建\u lst.append(lst[i]*lst[i+1]) 打印新列表

        5
  •  1
  •   BlackBeard    7 年前

    Number = 73167176531330624919225119674426574742355349194934
    
    print(max(int(x)*int(y) for x, y in zip(str(Number)[:-1], str(Number)[1:])))
    

    答案是54(9*6)

        6
  •  1
  •   jpp    7 年前

    不需要创建列表来计算iterable的最大值。这将是低效的。下面是一个函数解决方案,使用 max map mul :

    from operator import mul
    
    x = '73167176531330624919225119674426574742355349194934'
    
    res = max(map(mul, map(int, x), map(int, x[1:]))) # 54
    
        7
  •  1
  •   hiro protagonist    7 年前

    严格使用整数算术的版本(与字符串相反):

    Number = 73167176531330624919225119674426574742355349194934
    
    max_prod, a, b = 0, None, None
    
    Number, prev = divmod(Number, 10)
    while Number:
        Number, cur = divmod(Number, 10)
        prod = cur * prev
        if prod > max_prod:
            max_prod, a, b = prod, cur, prev
            # if prod == 81:  # prod will never be bigger; can as well stop here.
            #     break
        prev = cur
    
    print(a, b, max_prod)  # 9 6 54
    

    更新以显示这将如何适用于您当前的问题(您将查找13个相邻数字的最大乘积):

    Number = int('731671765313306249192251196744265747423553491949349698352031277450632623957831801698480186947885184385861'
                 '560789112949495459501737958331952853208805511125406987471585238630507156932909632952274430435576689664895'
                 '044524452316173185640309871112172238311362229893423380308135336276614282806444486645238749303589072962904'
                 '915604407723907138105158593079608667017242712188399879790879227492190169972088809377665727333001053367881'
                 '220235421809751254540594752243525849077116705560136048395864467063244157221553975369781797784617406495514'
                 '929086256932197846862248283972241375657056057490261407972968652414535100474821663704844031998900088952434'
                 '506585412275886668811642717147992444292823086346567481391912316282458617866458359124566529476545682848912'
                 '883142607690042242190226710556263211111093705442175069416589604080719840385096245544436298123098787992724'
                 '428490918884580156166097919133875499200524063689912560717606058861164671094050775410022569831552000559357'
                 '2972571636269561882670428252483600823257530420752963450')
    
    def digit_prod(i):
        '''
        return the product of all the decimal digits in i
        '''
        if i == 0:
            return 0
        prod = 1
        while i:
            i, r = divmod(i, 10)
            if r == 0:
                return 0
            prod *= r
        return prod
    
    
    max_prod, res = 0, None
    
    mod = 10**13
    while Number:
        i = Number % mod
        prod = digit_prod(Number % mod)
        if prod > max_prod:
            max_prod, res = prod, i
            # if prod == 9**13: # prod will never be bigger; can as well stop here.
            #     break
        Number //= 10
    
    print(res, max_prod)  # 5576689664895 23514624000
    
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