Python中成绩判断

Question

90一100优秀

80-90良好

70 -80中等

60-70及格

60以下不及格

根据用户输入的成绩，给出优良中及格不及格的评语

Answer 1

a = int(input())

if 90<= a and a <= 100:
    print("优秀")
elif 80<=a and a<90:
    print("良好")
elif 70 <= a and a<80:
    print("中等")
elif 60<=a and a<70:
    print("及格")
else:
    print("不及格")