题解 | #人民币转换#

这题单纯的条件比较多

num = input()
intpart, floatpart = num.split('.')  # 小数点前后分开处理
length = len(intpart)  # 整数部分长度

resStr = "人民币"
if num == 0:
    print("人民币零元整")
rmbDict = {1:'壹', 2:'贰', 3:'叁', 4:'肆', 5:'伍',
          6:'陆', 7:'柒', 8:'捌', 9:'玖'}
powDict = {1:'拾', 2:'佰',3:'仟'}
count = length  # 小数点前的位置

# *********** 先写小数点前 ***********
preZeroFlag = False  # 初始化0标记
while count > 0:
    currNumStr = num[length - count]  # 当前数字，字符形式
    currNum = int(currNumStr)
    if length == 1 and currNum == 0: # 若整数部分为0则只写小数部分
        break
    
    # 根据规则写字符串
    if currNum == 0:
        preZeroFlag = True
    elif currNum != 0:
        # 若前面有0则加入汉字零
        if preZeroFlag == True:
            if count == 4:  # 规则5：千前面有0不加零
                continue
            else:
                resStr += '零'
            preZeroFlag = False
        # 万位、亿位、个位元单独处理
        if count == 5:
            resStr += (rmbDict[currNum] + '万')  # 万
        elif count == 9:
            resStr += (rmbDict[currNum] + '亿')  # 亿
        elif count == 1:
            resStr += (rmbDict[currNum])  # 个位元
        else:
            # 取余
            if count > 8:
                    countMod = (count - 1) % 8
            else:
                countMod = (count - 1) % 4
            if currNum == 1 and countMod == 1:  # 规则4：10写作拾
                resStr += '拾'
            else:
                resStr += (rmbDict[currNum] + powDict[countMod])
    count -= 1
    if count == 0:  # 整数部分写完加汉字'元'
        resStr += '元'
        
# *********** 再写小数点后 ***********
if floatpart[0] == '0' and floatpart[1] == '0':
    resStr += '整'
elif floatpart[0] == '0' and floatpart[1] != '0':
    resStr += rmbDict[int(floatpart[1])] + '分'
elif floatpart[0] != '0' and floatpart[1] == '0':
    resStr += rmbDict[int(floatpart[0])] + '角'
else:
    resStr += rmbDict[int(floatpart[0])] + '角' + rmbDict[int(floatpart[1])] + '分'

print(resStr)