title:

The four adjacent digits in the 1000-digit number that have the greatest product are 9
9
8
9 = 5832.

73167176531330624919225119674426574742355349194934

96983520312774506326239578318016984801869478851843

85861560789112949495459501737958331952853208805511

12540698747158523863050715693290963295227443043557

66896648950445244523161731856403098711121722383113

62229893423380308135336276614282806444486645238749

30358907296290491560440772390713810515859307960866

70172427121883998797908792274921901699720888093776

65727333001053367881220235421809751254540594752243

52584907711670556013604839586446706324415722155397

53697817977846174064955149290862569321978468622482

83972241375657056057490261407972968652414535100474

82166370484403199890008895243450658541227588666881

16427171479924442928230863465674813919123162824586

17866458359124566529476545682848912883142607690042

24219022671055626321111109370544217506941658960408

07198403850962455444362981230987879927244284909188

84580156166097919133875499200524063689912560717606

05886116467109405077541002256983155200055935729725

71636269561882670428252483600823257530420752963450

Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?

翻译：

求出连续的13位数的积，使其最大

a="73167176531330624919225119674426574742355349194934"

a1="96983520312774506326239578318016984801869478851843"

a2="85861560789112949495459501737958331952853208805511"

a3="12540698747158523863050715693290963295227443043557"

a4="66896648950445244523161731856403098711121722383113"

a5="62229893423380308135336276614282806444486645238749"

a6="30358907296290491560440772390713810515859307960866"

a7="70172427121883998797908792274921901699720888093776"

a8="65727333001053367881220235421809751254540594752243"

a9="52584907711670556013604839586446706324415722155397"

a10="53697817977846174064955149290862569321978468622482"

a11="83972241375657056057490261407972968652414535100474"

a12="82166370484403199890008895243450658541227588666881"

a13="16427171479924442928230863465674813919123162824586"

a14="17866458359124566529476545682848912883142607690042"

a15="24219022671055626321111109370544217506941658960408"

a16="07198403850962455444362981230987879927244284909188"

a17="84580156166097919133875499200524063689912560717606"

a18="05886116467109405077541002256983155200055935729725"

a19="71636269561882670428252483600823257530420752963450"

a=a+a1+a2+a3+a4+a5+a6+a7+a8+a9+a10+a11+a12+a13+a14+a15+a16+a17+a18+a19

resu=0

for i in range(0,987):

	s=1

	for j in range(13):

		s*=int(a[i+j])

	if s > resu:

		resu = s

print resu

projecteuler---->problem=8----Largest product in a series的更多相关文章

  1. projecteuler Problem 8 Largest product in a series

    The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = ...

  2. Problem 8: Largest product in a series

    先粘实现代码,以后需要再慢慢补充思路 s = ''' 73167176531330624919225119674426574742355349194934 9698352031277450632623 ...

  3. Largest product in a series

    这个我开始理解错了,算错了. 我以为是求连续5个数的最大值,结果,是连接5个数相乘的最大值. 英语不好,容易吃亏啊. Find the greatest product of five consecu ...

  4. 【Project Euler 8】Largest product in a series

    题目要求是: The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × ...

  5. Project Euler 8 Largest product in a series

    题意:寻找这 1000 个数中相邻 13 个数相乘积最大的值 思路:首先想到的是暴力,但是还可以利用之前记录过的数值,什么意思呢?即在计算 2 - 14 后,再计算 3 - 15 时完全可以利用之前计 ...

  6. Largest product from 3 integers

    https://github.com/Premiumlab/Python-for-Algorithms--Data-Structures--and-Interviews/blob/master/Moc ...

  7. Problem 4: Largest palindrome product

    A palindromic number reads the same both ways. The largest palindrome made from the product of two 2 ...

  8. R语言学习——欧拉计划（11）Largest product in a grid

    Problem 11 In the 20×20 grid below, four numbers along a diagonal line have been marked in red. 08 0 ...

  9. Largest product in a grid

    这个比前面的要复杂点,但找对了规律,还是可以的. 我逻辑思维不强,只好画图来数数列的下标了. 分四次计算,存入最大值. 左右一次,上下一次,左斜一次,右斜一次. In the 2020 grid be ...

随机推荐

  1. FSM（有限状态机）

    游戏引擎是有限状态机最为成功的应用领域之一,由于设计良好的状态机能够被用来取代部分的人工智能算法,因此游戏中的每个角色或者器件都有可能内嵌一个状态机.考虑RPG游戏中城门这样一个简单对象,它具有打开( ...

  2. leetcode Reverse Integer python

    class Solution(object): def reverse(self, x): """ :type x: int :rtype: int "&quo ...

  3. CI(-)框架结构

    一 CI 是什么 CodeIgniter is an Application Development Framework - a toolkit - for people who build web ...

  4. 中国 省会 地级市 经纬度 city array

    <?php $city_arr = array ( '北京' => array ( 'gis_lng' => '116.405285', 'gis_lat' => '39.90 ...

  5. xmemcached的time out

    最近维护线上发现不停有java.util.concurrent.TimeoutException: Timed out(200) waiting for operation的问题,排查程序.配置文件的 ...

  6. ICT测试原理

    在线测试,ICT,In-Circuit Test,是通过对在线元器件的电性能及电气连接进行测试来检查生产制造缺陷及元器件不良的一种标准测试手段.它主要检查在线的单个元器件以及各电路网络的开.短路情况, ...

  7. Windows 7各版本的主要功能区别是什么 有何不同

    Windows 7包含6个版本,分别为Windows 7 Starter(初级版).Windows 7 Home Basic(家庭普通版).Windows 7 Home Premium(家庭高级版). ...

  8. 解决[warn] _default_ VirtualHost overlap on port 80, the first has precedence问题

    问题背景: 在apache的httpd.conf里新增加了1个VirtualHost,域名是xxx.com,此时,服务器总共2个VirtualHost ,service httpd restart的时 ...

  9. cocos2dx CCControlSwitch

    CCControlSwitch也是extension中的控件,本身比较简单,直接上例子 // on "init" you need to initialize your insta ...

  10. HDU 5782 Cycle（KMP+Hash）

    [题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=5782 [题目大意] 给出两个字符串,判断他们每一个前缀是否循环同构,循环同构的意思就是,字符串首位 ...