[已解决]Python：每日一题 332

zltzlt · 发表于 2020-2-14 19:28:25

William4869 发表于 2020-2-14 19:27
继续

可以了

28 ms

AINIDEREN · 发表于 2020-2-14 20:24:32

认真看老爷们答题

TJBEST · 发表于 2020-2-14 20:54:34

本帖最后由 TJBEST 于 2020-2-14 21:03 编辑

这题考验的是耐心和英文基本功

def fun332(num):
    def get3Num(string):
        num = int(string)
        if num == 0:
            return ''
        elif num>0 and num <10:
            return Dec[num]
        elif num>9 and num <20:
            return Teen[num - 10]
        elif num >19 and num <100:
            temp = TenTimes[num//10] + ' ' + get3Num(string[-1])
            if temp[-1] == ' ':
                return temp[:-1]
            else:
                return temp
        else:
            temp =  Dec[num//100] + ' '+ 'Hundred' + ' '+ get3Num(string[1:])
            if temp[-1] == ' ':
                return temp[:-1]
            else:
                return temp
    
    if num == 0:
        return 'Zero'
   
    Dec = ['Zero','One','Two','Three','Four','Five','Six','Seven','Eight','Nine']
    Teen = ['Ten','Eleven','Twelve','Thirteen','Fourteen','Fifteen','Sixteen','Seventeen','Eighteen','Nineteen']
    TenTimes = ['','','Twenty','Thirty','Forty','Fifty','Sixty','Seventy','Eighty','Ninety']
    Others = ['','Thousand','Million','Billion']

    string = str(num)
    M = len(string)
    
    seg = (M - 1) // 3
    op = M - seg * 3
    
    result = get3Num(string[0:op]) + ' ' + Others[seg]
    for index in range(0,seg):
        temp = get3Num(string[(op + 3* index):(op + 3*index + 3)])
        if temp != '':
            result += ' ' + temp + ' ' + Others[seg - 1 - index]
        else:
            pass
    if result[-1] == ' ':
        return result[:-1]
    else:
        return result

zltzlt · 发表于 2020-2-14 21:56:22

TJBEST 发表于 2020-2-14 20:54
这题考验的是耐心和英文基本功

36 ms

阴阳神万物主 · 发表于 2020-2-14 22:37:23

zltzlt 发表于 2020-2-14 18:41
不可行……

那么，百度翻译？

阴阳神万物主 · 发表于 2020-2-14 23:57:45

我就是要加 And 你来咬我呀~~

dic = {2:['','','Twenty','Thirty','Forty','Fifty','Sixty','Seventy','Eighty','Ninety'],
       3:['','One','Two','Three','Four','Five','Six','Seven','Eight','Nine'],
       1:['Ten','Eleven','Twelve','Thirteen','Fourteen','Fifteen','Sixteen',
        'Seventeen','Eighteen','Nineteen'],
       0:['','Thousand','Million','Billion']}
def solve(n):
    num = [int(x)for x in str(n)]
    temp = [num[(i-3 if i-3>=0 else 0):i] for i in range(len(num),0,-3)]
    res = ''
    for i in range(len(temp)):
        while len(temp[i])<3:
            temp[i].insert(0,0)
        trs = [dic[3][temp[i][0]]]
        if temp[i][1] == 1:
            trs.append(dic[1][temp[i][2]])
        else:
            trs.extend([dic[2][temp[i][1]],dic[3][temp[i][2]]])
        if trs[0]:
            if trs[1] or trs[2]:
                trs.insert(1,'And')
            trs.insert(1,'Hundred')
        elif (not i) and (trs[1] or trs[2]):
            trs.insert(1,'And')
        while '' in trs:
            trs.remove('')
        if trs:trs.append(dic[0][i])
        if res:
            res = ' '.join(trs)+' '+res
        else:
            res = ' '.join(trs)
    return res if res else 'Zero'
if __name__ == '__main__':
    print('示例1 输出：',repr(solve(123)))
    print('示例2 输出：',repr(solve(12345)))
    print('示例3 输出：',repr(solve(1234567)))
    print('示例4 输出：',repr(solve(1234567891)))

悲哀的猫 · 发表于 2020-2-14 23:58:54

我的天

悲哀的猫 · 发表于 2020-2-14 23:59:36

首先把它差分程一个个数字，然后再一一对应，我的天，太难做了

阴阳神万物主 · 发表于 2020-2-15 00:00:33

TJBEST 发表于 2020-2-14 18:20
楼主你这个不对举例

实例1：

示例3的小横线补上口牙！
百位的后面不是还要加 And 么？

ouyunfu · 发表于 2020-2-15 06:31:47

import locale;

NUMBER_CONSTANT = {0:"Zero ", 1:"One", 2:"Two", 3:"Three", 4:"Four", 5:"Five", 6:"Six", 7:"Seven",
            8:"Eight", 9:"Nine", 10:"Ten", 11:"Eleven", 12:"Twelve", 13:"Thirteen",
            14:"Fourteen", 15:"Fifteen", 16:"Sixteen", 17:"Seventeen", 18:"Eighteen", 19:"Nineteen" };
IN_HUNDRED_CONSTANT = {2:"Twenty", 3:"Thirty", 4:"Forty", 5:"Fifty", 6:"Sixty", 7:"Seventy", 8:"Eighty", 9:"Ninety"}
BASE_CONSTANT = {0:" ", 1:"Hundred", 2:"Thousand", 3:"Million", 4:"Billion"};
def F332(number):
if str(number).isnumeric():
      if str(number)[0] == '0' and len(str(number)) > 1:
         return translateNumberToEnglish(int(number[1:]));
      if int(number) < 20:
         return NUMBER_CONSTANT[int(number)];
      elif int(number) < 100:
         if str(number)[1] == '0':
            return IN_HUNDRED_CONSTANT[int(str(number)[0])];
         else:
            return IN_HUNDRED_CONSTANT[int(str(number)[0])] + " " + NUMBER_CONSTANT[int(str(number)[1])];
      else:
         locale.setlocale(locale.LC_ALL, "English_United States.1252");
         strNumber = locale.format_string("%d" , number, grouping=True);
         numberArray = str(strNumber).split(",");
         stringResult = "";
         groupCount = len(numberArray) + 1;
         for groupNumber in numberArray:
            if groupCount > 1 and groupNumber[0:] != "000":
                  stringResult += str(f(str(groupNumber))) + " ";
            else:
                  break;
            groupCount -= 1;
            if groupCount > 1:
                  stringResult += BASE_CONSTANT[groupCount] + " ";
         endPoint = len(stringResult) - len(" hundred");
         return stringResult;

else:
      print("please input a number!");

def f(number):
if str(number).isnumeric() and len(number) < 4:
      if len(number) < 3:
         return F332(int(number));
      elif len(number) == 3 and number[0:] == "000":
         return " ";
      elif len(number) == 3 and number[1:] == "00":
         return NUMBER_CONSTANT[int(number[0])] + "  " + BASE_CONSTANT[1];
      else:
         return NUMBER_CONSTANT[int(number[0])] + "  " + BASE_CONSTANT[1] + "  " + F332((number[1:]));
else:
      print("number must below 1000");
print(F332(123))
print(F332(12345))
print(F332(1234567))
print(F332(1234567891))
print(F332(2147483648))

nl2020 · 发表于 2020-2-15 09:27:14

kinkon · 发表于 2020-2-15 09:51:52

没有含小数吗？

白咕咕 · 发表于 2020-2-15 12:48:37

num = input('请输入数字进行翻译：')

while num:
    if num.isdigit == False:
        print('输入错误！请重新输入：')
    if int(num) >= 2147483648 or int(num) < 0:
        print('输入错误！请重新输入：')
    else:
        break

n_list = list(str(num))

def read_1(n_list):
    dict1 = {
        '1' : 'One',
        '2' : 'Two',
        '3' : 'Three',
        '4' : 'Four',
        '5' : 'Five',
        '6' : 'Six',
        '7' : 'Seven',
        '8' : 'Eight',
        '9' : 'Nine'
        }
    
    n1 = n_list[-1]
    return dict1[n1]

def read_2(n_list):
    dict2 = {
        '1' : 'Ten',
        '11' : 'Eleven',
        '12' : 'Twelve',
        '13' : 'Thirteen',
        '14' : 'Fourteen',
        '15' : 'Fifteen',
        '16' : 'Sixteen',
        '17' : 'Seventeen',
        '18' : 'Eighteen',
        '19' : 'Nineteen',
        '2' : 'Twenty',
        '3' : 'Thirty',
        '4' : 'Forty',
        '5' : 'Fifty',
        '6' : 'Sixty',
        '7' : 'Seventy',
        '8' : 'Sighty',
        '9' : 'Ninety'
    }
    n1 = n_list[-1]
    n2 = n_list[-2]
    number = n2 + n1
    if number in dict2 == True:
        result = dict2[number]
    else:
        result = dict2[n2] + ' ' + read_1(n1)
    return result
    
def read_3(n_list):
    n3 = n_list[-3]
    return read_1(n3) + ' Hundred ' + read_2(n_list[-2] + n_list[-1])

def read(n_list):
    l = len(n_list)
    num3 = n_list[l-3:]
    num6 = n_list[l-6:l-3]
    num9 = n_list[l-9:l-6]
    t = 'Thousand'
    m = 'Millon'
    b = 'Billon'
    if l == 1:
        if '0' in n_list:
            return '0'
        else:
            return read_1(n_list)
    elif l == 2:
        return read_2(n_list)
    elif l == 3:
        return read_3(n_list)
    elif l == 4:
        num3 = n_list[l-3:]
        return read_1(n_list[0]) +' '+ t + ' '+ read_3(num3)
    elif l == 5:
        return read_2(n_list[:2]) + ' '+ t + ' '+ read_3(num3)
    elif l == 6:
        return read_3(num6) + ' '+ t + ' '+ read_3(num3)
    elif l == 7:
        return read_1(n_list[0]) + ' '+ m + ' '+ read_3(num6) + ' '+ t + ' '+ read_3(num3)
    elif l == 8:
        return read_2(n_list[:2]) + ' '+ m + ' '+ read_3(num6) + ' '+ t + ' '+ read_3(num3) 
    elif l == 9:
        return read_3(num9) + ' '+ m + ' '+ read_3(num6) + ' '+ t + ' '+ read_3(num3) 
    elif l == 10:
        return read_1(n_list[0]) + ' '+ b + ' '+ read_3(num9) + ' '+ m + ' '+ read_3(num6) + ' '+ t + ' '+ read_3(num3)

english = read(n_list)
print('翻译结果是：')
print(english)

楼主看这个行不行

scanhack · 发表于 2020-2-15 15:09:05

最后的魁拔发表于 2020-2-14 18:49
为啥这个就可以呢

百度可以哦

陈尚涵 · 发表于 2020-2-15 16:51:39

做一个叫'单位'的列表，在做一个叫'数字'的列表，接着做一个函数定义确认不就好了

？

zltzlt · 发表于 2020-2-15 16:55:34

阴阳神万物主发表于 2020-2-14 23:57
我就是要加 And 你来咬我呀~~

输入 1000010，多了一个空格

zltzlt · 发表于 2020-2-15 16:56:23

塔利班发表于 2020-2-14 18:47
101怎么整One Hundred And One ？要加And的么

不需要，加也行

zltzlt · 发表于 2020-2-15 16:58:20

ouyunfu 发表于 2020-2-15 06:31
import locale;

NUMBER_CONSTANT = {0:"Zero ", 1:"One", 2:"Two", 3:"Three", 4:"Four", 5:"Five", 6: ...

locale.Error: unsupported locale setting

zltzlt · 发表于 2020-2-15 17:01:29

白咕咕发表于 2020-2-15 12:48
楼主看这个行不行

输入 12345，结果有误

阴阳神万物主 · 发表于 2020-2-15 18:41:07

zltzlt 发表于 2020-2-15 16:55
输入 1000010，多了一个空格

这样应该就没有多的空格了。

dic = {2:['','','Twenty','Thirty','Forty','Fifty','Sixty','Seventy','Eighty','Ninety'],
       3:['','One','Two','Three','Four','Five','Six','Seven','Eight','Nine'],
       1:['Ten','Eleven','Twelve','Thirteen','Fourteen','Fifteen','Sixteen',
        'Seventeen','Eighteen','Nineteen'],
       0:['','Thousand','Million','Billion']}
def solve(n):
    num = [int(x)for x in str(n)]
    temp = [num[(i-3 if i-3>=0 else 0):i] for i in range(len(num),0,-3)]
    res = ''
    for i in range(len(temp)):
        while len(temp[i])<3:
            temp[i].insert(0,0)
        trs = [dic[3][temp[i][0]]]
        if temp[i][1] == 1:
            trs.append(dic[1][temp[i][2]])
        else:
            trs.extend([dic[2][temp[i][1]],dic[3][temp[i][2]]])
        if trs[0]:
            if trs[1] or trs[2]:
                trs.insert(1,'And')
            trs.insert(1,'Hundred')
        elif (not i) and (trs[1] or trs[2]):
            trs.insert(1,'And')
        while '' in trs:
            trs.remove('')
        if trs:
            if i:trs.append(dic[0][i])
            if res:
                res = ' '.join(trs)+' '+res
            else:
                res = ' '.join(trs)
    return res if res else 'Zero'
if __name__ == '__main__':
    print('示例1 输出：',repr(solve(123)))
    print('示例2 输出：',repr(solve(12345)))
    print('示例3 输出：',repr(solve(1234567)))
    print('示例4 输出：',repr(solve(1234567891)))
    print('之前多空格的 输出：',repr(solve(1000010)))

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[已解决]Python：每日一题 332

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