修改了一版,这次考虑得比较全,只是验证a, b, c的合法性的方法还不是很好,当有两个以上的数不合法时每次只报错一个,大家有其它好方法的麻烦告知一下
# -*- coding: utf-8 -*- import math a = float(input('please input a = ')) b = float(input('please input b = ')) c = float(input('please input c = ')) def quadratic(a, b, c): if not isinstance (a, (int, float)): raise TypeError ('error aaaaaa') if not isinstance (b, (int, float)): raise TypeError ('error bbbbbb') if not isinstance (c, (int, float)): raise TypeError ('error cccccc') delta = b*b-4*a*c if a == 0: x1 = -c/b return x1 elif (delta) > 0: x1 = (-b + math.sqrt(delta)) / (2*a) x2 = (-b - math.sqrt(delta)) / (2*a) return x1, x2 elif (delta) == 0: x1 = -b / (2*a) return x1 else: print ("无实数解") print(quadratic(a, b, c))
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yayayaxs
修改了一版,这次考虑得比较全,只是验证a, b, c的合法性的方法还不是很好,当有两个以上的数不合法时每次只报错一个,大家有其它好方法的麻烦告知一下