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15-avocado-13 #71

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Feb 21, 2024
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15-avocado-13 #71

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11 changes: 11 additions & 0 deletions avocado-13/λ‹€μ΄λ‚˜λ―Ή ν”„λ‘œκ·Έλž˜λ°/1둜 λ§Œλ“€κΈ°.py
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n = int(input())
d = [0] * (n + 1) ## d에 κ³„μ‚°λœ 값을 μ €μž₯ν•΄λ‘”λ‹€. n + 1이라고 ν•œ μ΄μœ λŠ”, 1번째 μˆ˜λŠ” 사싀 d[1]이 μ•„λ‹ˆκ³  d[2]이기 λ•Œλ¬Έμ—, κ³„μ‚°ν•˜κΈ° νŽΈν•˜κ²Œ d[1]을 1번째 인 것 처럼 λ§Œλ“€μ–΄μ€€λ‹€.

for i in range(2, n + 1):

d[i] = d[i - 1] + 1
if i % 3 == 0:
d[i] = min(d[i], d[i // 3] + 1)
if i % 2 == 0:
d[i] = min(d[i], d[i // 2] + 1)
print(d[n])
Comment on lines +4 to +11
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저도 μ‹œν˜„λ‹˜κ³Ό 같은 νλ¦„μœΌλ‘œ 해결을 ν–ˆλ„€μš” γ…Žγ…Ž

n = int(input())
dp = [0] * (10**6+1) 

for i in range(2,n+1):
    # 1을 i번 λΉΌλŠ” 경우의 μˆ˜λž‘
    dp[i] = dp[i-1] + 1
    # 2의 배수면 2둜 λ‚˜λˆ„λŠ” 경우의 μˆ˜μ™€ λΉ„κ΅ν•΄μ„œ 적은 κ°’
    if not i%2:
        dp[i] = min(dp[i], dp[i//2]+1)
    # 3의 배수면 3으둜 λ‚˜λˆ„λŠ” 경우의 μˆ˜μ™€ λΉ„κ΅ν•΄μ„œ 적은 κ°’
    if not i%3:
        dp[i] = min(dp[i], dp[i//3]+1)
print(dp[n])

μ²˜μŒμ— i-1의 1둜 λͺ¨λ‘ λΉΌμ£ΌλŠ” 경우의 μˆ˜μ™€ ν•˜λ‚˜μ˜ 쀄 μ•ˆμ—μ„œ 비ꡐλ₯Ό ν•˜λ €λ‹€ μ‹€νŒ¨ν–ˆλŠ”λ°
λ‹€λ₯Έ μ‚¬λžŒλ“€μ˜ μ½”λ“œλ₯Ό 보며 μ–΄λ–»κ²Œ κ΅¬ν˜„ν•΄μ•Όν• μ§€λ₯Ό μ•Œκ²Œ λ˜μ—ˆμŠ΅λ‹ˆλ‹€ :)

2λ‚˜ 3으둜 λ‚˜λˆ μ§€λŠ” μ΅œμ†Œ νšŸμˆ˜μ— 숫자둜 λ‚˜λˆˆ λ‚˜λ¨Έμ§€ 만큼 (2 or 3) 1을 λΉΌμ€˜μ•Όν•˜λŠ” 경우의 수λ₯Ό μΆ”κ°€ν•œ 것을
ν•œκΊΌλ²ˆμ— λΉ„κ΅ν•˜μ—¬ μ΅œμ†Ÿκ°’μ„ 찾아쀄 수 μžˆμ—ˆμŠ΅λ‹ˆλ‹€ ✨

for i in range(2,N+1):
    dp[i] = min(dp[i-1]+1, dp[i//2]+1 + i%2, dp[i//3]+1 + i%3)
print(A[N])

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@avocado-13 avocado-13 Feb 20, 2024 •
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μš°μ™€ 2의 λ°°μˆ˜μΈμ§€ μ•„λ‹Œμ§€ ν™•μΈν•˜λŠ” 쑰건을 if not i%2 둜 μ λŠ” 것은 처음 λ΄€μ–΄μš” (βœΏβ—•β€Ώβ—•βœΏ) ν•œκΊΌλ²ˆμ— κ΅¬ν•˜λŠ” 것도 ... bb