本题最标准做法:(亲码亲测可用)
import math
def quadratic(a,b,c): if not isinstance(a, (int, float)) & isinstance(b, (int, float)) & isinstance(c, (int, float)) : raise TypeError('bad operand type')
delta=(bb-4ac) if delta>0: x1=(-b+math.sqrt(delta))/(2a) x2=(-b-math.sqrt(delta))/(2a) return x1,x2 elif delta==0: x=(-b+sqrt(delta))/2a return x else: return
print('quadratic(2, 3, 1) =', quadratic(2, 3, 1)) print('quadratic(1, 3, -4) =', quadratic(1, 3, -4))
if quadratic(2, 3, 1) != (-0.5, -1.0): print('测试失败') elif quadratic(1, 3, -4) != (1.0, -4.0): print('测试失败') else: print('测试成功')
#但是你们发现了吗? 单这个题来讲...
↓这玩意就够了的说
gd = math.sqrt(bb-4ac) y1 = (-b+gd)/(2a) y2 = (-b-gd)/(2*a) return y1, y2
#......emmmmmm
特注:网页吃掉了源代码中所有的 * 号,粘贴时请记得手动添加
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troublesaint
本题最标准做法:(亲码亲测可用)
-- coding: utf-8 --
import math
def quadratic(a,b,c): if not isinstance(a, (int, float)) & isinstance(b, (int, float)) & isinstance(c, (int, float)) : raise TypeError('bad operand type')
delta=(bb-4ac) if delta>0: x1=(-b+math.sqrt(delta))/(2a) x2=(-b-math.sqrt(delta))/(2a)
return x1,x2 elif delta==0: x=(-b+sqrt(delta))/2a return x else: return
print('quadratic(2, 3, 1) =', quadratic(2, 3, 1)) print('quadratic(1, 3, -4) =', quadratic(1, 3, -4))
if quadratic(2, 3, 1) != (-0.5, -1.0): print('测试失败') elif quadratic(1, 3, -4) != (1.0, -4.0): print('测试失败') else: print('测试成功')
#但是你们发现了吗? 单这个题来讲...
gd = math.sqrt(bb-4ac) y1 = (-b+gd)/(2a) y2 = (-b-gd)/(2*a) return y1, y2
#......emmmmmm
特注:网页吃掉了源代码中所有的 * 号,粘贴时请记得手动添加