This is an example of the answer to the 2016 summer hospital exam. Problem link
--N-ary number
(1)
parse1obj = {
'0' : 0,
'1' : 1,
'2' : 2,
'3' : 3
}
def FtoT(f):
f = str(f)
ret = 0
for i in range(0, len(f)):
idx = len(f) - i - 1
ret += parse1obj[f[idx]] * (4**(i))
return ret
(2)
parse2obj = {
'a': 0,
'b': 1,
'c': 2,
'd': 3,
'e': 4,
'f': 5,
'g': 6,
'h': 7
}
def EtoT(e):
ret = 0
for i in range(0, len(e)):
idx = len(e) - i - 1
ret += parse2obj[e[idx]] * (8**(i))
return ret
(3) MMXV
(4)
dict4 = {
'I': 1,
'V': 5,
'X': 10,
'L': 50,
'C': 100,
'D': 500,
'M': 1000
}
dict4_1 = {
'IV': 4,
'IX': 9,
'XL': 40,
'XC': 90,
'CD': 400,
'CM': 900
}
# ruturn bool
def is_reverse(cur_ch, next_ch):
if (cur_ch == 'I'):
return next_ch == 'V' or next_ch == 'X'
if (cur_ch == 'X'):
return next_ch == 'L' or next_ch == 'C'
if (cur_ch == 'C'):
return next_ch == 'D' or next_ch == 'M'
return False
def parser4(rome):
ret = 0
i = 0
while (i < len(rome)):
cur_ch = rome[i]
if (i < len(rome) - 1):
next_ch = rome[i + 1]
if (is_reverse(cur_ch, next_ch)):
key_str = cur_ch + next_ch
ret += dict4_1[key_str]
i += 2
else:
ret += dict4[cur_ch]
i += 1
else:
ret += dict4[cur_ch]
i += 1
return ret
(5)
def sub_parser5(num, digit):
if (digit == 4):
return 'M' * num
elif (digit == 3):
if (num == 9):
return 'CM'
elif (num == 4):
return 'CD'
elif (num >= 5 and num <= 8):
return 'D' + 'C' * (num - 5)
else:
return 'C' * num
elif (digit == 2):
if (num == 9):
return 'XC'
elif (num == 4):
return 'XL'
elif (num >= 5 and num <= 8):
return 'L' + 'X' * (num - 5)
else:
return 'X' * num
elif (digit == 1):
if (num == 9):
return 'IX'
elif (num == 4):
return 'IV'
elif (num >= 5 and num <= 8):
return 'V' + 'I' * (num - 5)
else:
return 'I' * num
def parser5(t):
str_t = str(t)
for i in range(len(str_t), 4):
str_t = '0' + str_t
ret = ''
for i in range(0, len(str_t)):
num = int(str_t[i])
ret += sub_parser5(num, len(str_t) - i)
return ret
(6)
def parser6(t):
str_t = str(t)
for i in range(len(str_t), 4):
str_t = '0' + str_t
str_t = str_t[::-1]
fir_num = int(str_t[0])
sec_num = int(str_t[1])
thr_num = int(str_t[2])
for_num = int(str_t[3])
if (fir_num == 9):
if (sec_num == 9):
if (thr_num == 9):
return (for_num * 'M' + 'IM')
elif (thr_num == 4):
return (for_num * 'M' + 'ID')
else:
return sub_parser5(for_num, 4) + sub_parser5(thr_num, 4) + 'IC'
elif (sec_num == 4):
return sub_parser5(for_num, 4) + sub_parser5(thr_num, 3) + 'IL'
else:
return sub_parser5(for_nu, 4) + sub_parser5(thr_num, 3) + sub_parser5(sec_num, 2) + 'IX'
else:
return parser5(t)
(7) (6) I think it's easier ... Since it is less than 100,000, both hundred and thousand will only appear once at the maximum. Use'zero'as a sentinel because it would be a problem if there was no second half after splitting
dict7 = {
'zero': 0,
'one': 1,
'two': 2,
'three': 3,
'four': 4,
'five': 5,
'six': 6,
'seven': 7,
'eight': 8,
'nine': 9,
'ten': 10,
'eleven': 11,
'twelve': 12,
'thirteen': 13,
'fourteen': 14,
'fifteen': 15,
'sixteen': 16,
'seventeen': 17,
'eighteen': 18,
'nineteen': 19,
'twenty': 20,
'thirty': 30,
'forty': 40,
'fifty': 50,
'sixty': 60,
'seventy': 70,
'eighty': 80,
'ninety': 90,
'hundred': 100,
'thousand': 1000,
}
def parser7(s):
#Add sentinel
s += ' zero'
ret = 0
array_s = s.split()
if ('thousand' in array_s):
t_s = ' '.join(array_s)
t_s = t_s.split('thousand')
t_0 = t_s[0]
t_s0 = t_0.split()
t_1 = t_s[1]
t_s1 = t_1.split()
tmp1 = 0
for i in range(0, len(t_s0)):
word = t_s0[i]
tmp1 += dict7[word]
ret += tmp1 * 1000
if ('hundred' in t_s1):
h_s = t_1.split('hundred')
h_0 = h_s[0]
h_s0 = h_0.split()
h_1 = h_s[1]
h_s1 = h_1.split()
tmp2 = 0
for j in range(0, len(h_s0)):
word = h_s0[j]
tmp2 += dict7[word]
ret += tmp2 * 100
for k in range(0, len(h_s1)):
word = h_s1[k]
ret += dict7[word]
else:
for j in range(0, len(t_s1)):
word = t_s1[j]
ret += dict7[word]
else:
for i in range(0, len(array_s)):
word = array_s[i]
if (dict7[word] == 100):
ret *= 100
else:
ret += dict7[word]
return ret
It takes more time to type than to think I think it's a year where you can complete the answer in time if you do it carefully
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