Project Euler - Problem 8
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Find the greatest product of five consecutive digits in the 1000-digit number.
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
从这1000位数字中,找出乘积最大的连续的5个数字。
#!/usr/bin/env python# coding: utf-8def ConsecDigitsProduct(s, n): p = 0 m = 0 for i in range(len(s)-4): p = int(s[i])*int(s[i+1])*int(s[i+2])*int(s[i+3])*int(s[i+4]); if p > m: m = p return mif __name__ == "__main__": # 把提供的大数当作字符串传入即可 print ConsecDigitsProduct('731...963450', 5)
其实如果连续的5个数字中有一个为0, 则周围的8个数字都不用考虑,可以跳过。
def LargestProduct(s, n): i = p = m = 0 while True: for k in range(n): if int(s[i+k]) == 0: i += n + k - 1 break if i > len(s) - n: break p = int(s[i])*int(s[i+1])*int(s[i+2])*int(s[i+3])*int(s[i+4]); if p > m: m = p i += 1 return m
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