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Merge pull request #257 from Suxxxxhyun/Suxxxxhyun
[Failed] PRG_피로도(복습필요), PRG_n^2 배열 자르기(복습필요), [Failed] PRG_k진수에서 소수 개수 구하기(복습필요)
2 parents 56b61bf + cdcad0e commit a3ba2ff

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# 문자열 뒤집기
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# 문자열[::-1]
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import math
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def isPrime(x):
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if int(x) == 1:
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return False
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else:
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for i in range(2, int(math.sqrt(int(x))) + 1):
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if int(x) % i == 0:
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return False
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return True
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def k_num(n,k):
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if k == 10:
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return n
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else:
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new_n = ''
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while n > 0:
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new_n += str(n % k)
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n //= k
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return new_n[::-1]
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def solution(n, k):
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answer = 0
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num = k_num(n,k)
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# print(num)
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nums = str(num).split("0")
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print(nums)
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for x in nums:
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if x and isPrime(x):
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answer += 1
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return answer
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print(solution(437674, 3))

Week6/공통/k진수에서_소수개수_구하기/test.java

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def solution(n, left, right):
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answer = []
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for i in range(left, right+1):
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answer.append(max((i // n), (i % n)) + 1)
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return answer
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print(solution(3, 2, 5))

Week6/공통/피로도/suhyun.py

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# 내장함수 이용해서 해보자. -> permutations()
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# from itertools import permutations
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# repeat을 쓰는 것은 product뿐임!
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# 8!
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# 던전을 최대로 탐험하고 싶음.
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from itertools import permutations
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import math
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answer = 0
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d_len = 0
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def solution(k, dungeons):
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global answer
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selected = [0] * len(dungeons) # 던전을 탐험하는 순서
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d_len = len(dungeons)
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for a in permutations(range(d_len), d_len):
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cnt = 0
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pe = k
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# print(a)
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for p in a:
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if dungeons[p][0] <= pe:
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pe -= dungeons[p][1]
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cnt += 1
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if answer < cnt:
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answer = cnt
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return answer
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print(solution(80, [[80,20],[50,40],[30,10]]))

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