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kot_mapku3

rome

Oct 21st, 2022
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Python 1.20 KB | None | 0 0
  1. # Определить структуру соответствия римской цифры и арабской
  2. # Считываем число (< 4000)
  3. # Разбиваем число на слагаемые, сохранем в список
  4. # Каждое слагаемое прогоняем по списку
  5. # Собираем ответ
  6.  
  7.  
  8. # 1000 -> M
  9. # 10 -> "IX"
  10.  
  11. # "X" -> 10, "IX" -> 10 - 1
  12. # "IV" -> 5 - 1
  13.  
  14. # "I", "V", "X", "L", "C", "D", "M"
  15. #  1,   5,   10,  50, 100, 500, 1000
  16.  
  17. # "XLIV" <- 44
  18. # "LV" <- 55
  19.  
  20. # [{1: "I"}, {5: "V"} ... ]
  21. # [["I", 1], ["V", 5] ... ]
  22.  
  23. # 50, 5
  24.  
  25. # L, V
  26.  
  27. # 65 -> LXV -> 50, 10, 5
  28. # 85 -> LXXXV -> 50, 10, 10, 10, 5
  29. # 95 -> XCV -> 100, 10, 5
  30.  
  31. # IC -> wrong 99 -> XCIX
  32.  
  33. # IX, IV
  34. # XC, XL
  35. # CM, CD
  36.  
  37.  
  38. # people_1 = {
  39. #     "alex": 23, "jack": 45
  40. # }
  41.  
  42. # people_1.update({"alex": 25})
  43.  
  44. # print(people_1.keys())
  45.  
  46. # print(people_1.values())
  47.  
  48. # people_2 = [["alex", 23], ["jack", 45]]
  49.  
  50. # for people in people_2:
  51. #     if people[0] == "alex":
  52. #         print("YES")
  53. #         break
  54.  
  55. # if "alex" in people_1:
  56. #     print("YES")
  57.  
  58. # people_1["alex"] # b4g5h673jdfkjd -> 75: alex -> 23
  59.  
  60. # # 1 sec. N -> N sec
  61.  
  62. # # N/2 sec. N * N/2 -> N^2/2
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