我將使用基本運算符制作一個 python 程序來請求用戶的收入，以便計算 2017 年和 2018 年的累進稅。輸出應如下所示：Income: 150002017 tax: (tax amount here) #will skip the math for samples2018 tax: (tax amount here)我的程序現在為 2017 / 2018 生成兩個打印，但將停止在 2018 嵌套 if 括號 82501 到 157500（嵌套在第 4 個 elif 中）......這是我的程序截至目前，因為它很長我會標出它停止工作的地方。income = float(input("Enter your income to calculate tax: "))#Loop input/calculationswhile income > 0:    print("----------------------------------------------")    if income >= 0 and income <= 9325:        bracket1 = income * 0.10        tax2017 = bracket1        print("Income: ",income)        print("2017 tax: ",format(tax2017,'.2f'))        if income >= 0 and income <= 9525:            newbracket1 = income * 0.10            tax2018 = newbracket1            print("2018 tax: ",format(tax2018,'.2f'))        income = float(input("\nEnter your income as and integer with no commas: "))    elif income >= 9326 and income <= 37950:        bracket1 = 9325 * 0.10        bracket2 = (income - 9326) * 0.15        tax2017 = bracket1 + bracket2        print("Income: ",income)        print("2017 tax: ",format(tax2017,'.2f'))        if income >= 9526 and income <=38700:            newbracket1 = 9526 * 0.10            newbracket2 = (income - 9525) * 0.12            tax2018 = newbracket1 + newbracket2            print("2018 tax: ",format(tax2018,'.2f'))        income = float(input("\nEnter your income as and integer with no commas: "))    elif income >= 37951 and income <= 91900:        bracket1 = 9325 * 0.10        bracket2 = (37950 - 9325) * 0.15        bracket3 = (income - 37951) * 0.25        tax2017 = bracket1 + bracket2 + bracket3        print("Income: ",income)我已經標記了行不通的行。只是為了澄清，嵌套的 if 和打印之前的輸出 2017 年和 2018 年的結果，但當收入在標記范圍內或更大時，只會打印 2017 年的稅。