【每日一题20220223】第几世纪

:woman_mage: 给定一个四位数组成的字符串表示年份year,请编写一个函数solution,返回它代表的是第几世纪标记,由数字和小写英文字母缩写组成。

备注:世纪末位数字是1则英文缩写成st(first),是2则缩写成nd(second),是3则缩写成rd(third),其他则缩写成(th)

示例:
输入: “1999”,输出:“20th”。
输入:“2022”,输出:“21st”。

题目难度:简单
题目来源:CodeWars:What century is it?

def solution(year: str) -> str:
    # your code here

assert solution("2022") == "21st"
assert solution("2154") == "22nd"
assert solution("2259") == "23rd"
assert solution("1234") == "13th"
def solution(year: str) -> str:
    # your code here
    m = int(year) // 100
    if (m+1) % 10 == 1:
        return "".join(str(m+1)+'st')
    elif (m+1) % 10 == 2:
        return "".join(str(m+1)+'nd')
    elif (m+1) % 10 == 3:
        return "".join(str(m+1)+'rd')
    else:return ''.join(str(m+1)+'th')

assert solution("2022") == "21st"
assert solution("2154") == "22nd"
assert solution("2259") == "23rd"
assert solution("1234") == "13rd"
assert solution('1456') == "15th"
def solution(year: str) -> str:
    num_base = int(int(year) / 100)
    digit_num = int(str(num_base)[-1])
    case_num = int(str(num_base)[0])
    y = {
        0 : str(num_base + 1) + "st",
        1 : str(num_base + 1) + "nd",
        2 : str(num_base + 1) + "rd",
    }
    if case_num >= 2 and digit_num in y.keys():
        return y.get(digit_num)
    else:
        return str(num_base + 1) + "th"

assert solution("2022") == "21st"
assert solution("2154") == "22nd"
assert solution("2259") == "23rd"
assert solution("1234") == "13th"

11-19 应该都是th结尾的

同上,13-19 都是th,因为11-19 的英文不是通过几十+几说的,是eleven、twelve、thirteen这种有独立单词的。从20起,之后的英文都是几十+几,例如21 是twenty+one,所以第21是21st

2 Likes

多谢大佬指出问题,我还以为最后一个断言是故意加的断言失败的情况 :sob:

def solution(year: str) -> str:
    shiji = int(year)//100 + 1
    if shiji >= 20:
        if str(shiji)[-1] == "1":
            return str(shiji)+"st"
        elif str(shiji)[-1] == "2":
            return str(shiji) + "nd"
        else:
            return str(shiji) + "rd"
    else:
        return str(shiji) + "th"

坑在英语单词上 :joy:


def solution(year: str) -> str:
    cen = int(year[:2]) + 1
    
    if cen % 10 == 1:   th = 'st'
    elif cen % 10 == 2: th = 'nd'
    elif cen % 10 == 3: th = 'rd'
    else:               th = 'th'
    
    if cen // 10 == 1:
        th = 'th'
    return str(cen) + th


assert solution("2022") == "21st"
assert solution("2154") == "22nd"
assert solution("2259") == "23rd"
assert solution("1234") == "13th"

def solution(year: str) -> str:
    y = int(year[0:2]) + 1
    if int(year[0]) >= 2:
        if y % 10 == 1:
            return str(y) + 'st'
        elif y % 10 == 2:
            return str(y) + 'nd'
        elif y % 10 == 3:
            return str(y) + 'rd'
        else:
            return str(y) + 'th'
    else:
        return str(y) + 'th'

assert solution("2022") == "21st"
assert solution("2154") == "22nd"
assert solution("2259") == "23rd"
assert solution("1234") == "13th"
def solution(year: str) -> str:
    part1 = int(int(year) / 100) # 取整
    part2 = int(year) % 100   # 取余
    if part2 != 0:
        part1 += 1
        if str(part1)[-1] == '1':
            return str(part1)+'st'
        elif str(part1)[-1] == '2':
            return str(part1) + 'nd'
        elif str(part1)[-1] == '3':
            return str(part1) + 'rd'
        else:
            return str(part1) + 'nd'

assert solution("2022") == "21st"
assert solution("2154") == "22nd"
assert solution("2259") == "23rd"
assert solution("1234") == "13rd"

:joy: 参考大佬们的说法,补上了11-19的逻辑

def solution(year: str) -> str:
    # your code here
    century_num = int(int(year)/100) + 1
    last_num = century_num % 10
    if last_num == 1:
        abbr_word = "st"
    elif last_num == 2:
        abbr_word = "nd"
    elif last_num == 3:
        abbr_word = "rd"
    else:
        abbr_word = "th"
    if century_num // 10 == 1:
        abbr_word = 'th'
    return str(century_num) + abbr_word

assert solution("2022") == "21st"
assert solution("2154") == "22nd"
assert solution("2259") == "23rd"
assert solution("1234") == "13th"

def solution(year):
    datas = {1:'st', 2:'nd', 3:'rd'}
    end = int(year[-2:])
    century = int(year[:2])

    if end > 0:
        century = century + 1
    else:
        century = century

    results = ''
    th_case = [x for x in range(11,20)]
    if century not in th_case and int(str(century)[-1])in datas:
        results = '%s%s' % (century, datas[int(str(century)[-1])])
    else:
        results = '%s%s' % (century, 'th')
    return results

def solution(year: str) -> str:
    shiji=str(int(year[:2])+1)
    if int(shiji)>=20:
        if shiji[-1]=='1':
            return shiji+'st'
        elif shiji[-1]=='2':
            return shiji + 'nd'
        elif shiji[-1]=='3':
            return shiji + 'rd'
    else:
        return shiji + 'th'
        def solution(year: str) -> str:
        century = int(year) // 100 + 1
        if 3 < century < 20:return str(century) + 'th'  #看了评论,有大佬,加了这句
        suffix = 'st' if century % 10 == 1 else 'nd' if century % 10 == 2 else 'rd' if century % 10 == 3 else 'th'
        return str(century) + suffix

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