Project Euler #8

Project Euler #8

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• 문제
  
  • 풀이
• 후기

문제

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The four adjacent digits in the 1000-digit number that have the greatest product are $9 \times 9 \times 8 \times 9 = 5832$.
$$> 73167176531330624919225119674426574742355349194934 \\ > 96983520312774506326239578318016984801869478851843 \\ > 85861560789112949495459501737958331952853208805511 \\ > 12540698747158523863050715693290963295227443043557 \\ > 66896648950445244523161731856403098711121722383113 \\ > 62229893423380308135336276614282806444486645238749 \\ > 30358907296290491560440772390713810515859307960866 \\ > 70172427121883998797908792274921901699720888093776 \\ > 65727333001053367881220235421809751254540594752243 \\ > 52584907711670556013604839586446706324415722155397 \\ > 53697817977846174064955149290862569321978468622482 \\ > 83972241375657056057490261407972968652414535100474 \\ > 821663704844031{\color{RED}{9989}}0008895243450658541227588666881 \\ > 16427171479924442928230863465674813919123162824586 \\ > 17866458359124566529476545682848912883142607690042 \\ > 24219022671055626321111109370544217506941658960408 \\ > 07198403850962455444362981230987879927244284909188 \\ > 84580156166097919133875499200524063689912560717606 \\ > 05886116467109405077541002256983155200055935729725 \\ > 71636269561882670428252483600823257530420752963450 >$$
Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?

다음 1000 자릿수에서 연속된 네 수의 곱 중 가장 큰 값은 $9 \times 9 \times 8 \times 9 = 5832$이다.
1000 자릿수에서 연속된 열세 수의 곱 중 가장 큰 값은 얼마인가?

풀이

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• while과 for로 루핑을 돌렸다.(노가다)

import time
num = 5
rs1 = 1
rs2 = 1
rl = []
i = 0

nl =  list('73167176531330624919225119674426574742355349194934')
nl += list('96983520312774506326239578318016984801869478851843')
nl += list('85861560789112949495459501737958331952853208805511')
nl += list('12540698747158523863050715693290963295227443043557')
nl += list('66896648950445244523161731856403098711121722383113')
nl += list('62229893423380308135336276614282806444486645238749')
nl += list('30358907296290491560440772390713810515859307960866')
nl += list('70172427121883998797908792274921901699720888093776')
nl += list('65727333001053367881220235421809751254540594752243')
nl += list('52584907711670556013604839586446706324415722155397')
nl += list('53697817977846174064955149290862569321978468622482')
nl += list('83972241375657056057490261407972968652414535100474')
nl += list('82166370484403199890008895243450658541227588666881')
nl += list('16427171479924442928230863465674813919123162824586')
nl += list('17866458359124566529476545682848912883142607690042')
nl += list('24219022671055626321111109370544217506941658960408')
nl += list('07198403850962455444362981230987879927244284909188')
nl += list('84580156166097919133875499200524063689912560717606')
nl += list('05886116467109405077541002256983155200055935729725')
nl += list('71636269561882670428252483600823257530420752963450')

stime = time.time()

while i + num <= 1000:
    for j in nl[i:i + num]:
        rs1 *= int(j)
    if rs1 > rs2:
        rs2 = rs1
        rl = nl[i:i + num]
    rs1 = 1
    i += 1

print(rl, rs2)
print(time.time() - stime)

후기

딱히 수학적인 지식을 요하거나 하는 부분은 없는 문제같다.