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從用戶處獲得的 2 個(gè)數(shù)字與 x、y、z 之和的乘積之差 (Python)

從用戶處獲得的 2 個(gè)數(shù)字與 x、y、z 之和的乘積之差 (Python)

慕碼人2483693 2023-08-08 10:29:30
x,y,z = 2,5,10number1 = int(input(""))number2 = int(input(""))Multiplication of numbers =  int(number1) * int(number2)print("Multiplication of numbers: " + str(Multiplication of numbers))x_y_z = int(x) + int(y) + int(z)x_y_z_toplam_ = int(Multiplication of numbers) + int(x_y_z)print(x_y_z_toplam_ )fark? = int(x_y_z_toplam_ ) - int(x_y_z_toplam)print(fark?)
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1 回答

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肥皂起泡泡

TA貢獻(xiàn)1829條經(jīng)驗(yàn) 獲得超6個(gè)贊

這是一個(gè)經(jīng)過(guò)清理和解釋的版本：


x,y,z = 2,5,10


number1 = int(input("Enter number 1: "))  # input prompt message


number2 = int(input("Enter number 2: "))


multiplication_of_numbers =  number1 * number2  # variable name cannot be with spaces, also no need to convert an already int type to an int


print("Multiplication of numbers: {}".format(multiplication_of_numbers))


x_y_z = x + y + z


x_y_z_toplam = multiplication_of_numbers + x_y_z


print(x_y_z_toplam)


fark? = x_y_z_toplam - x_y_z_toplam


print(fark?)


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