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一元二次方程求解都不会了。。。

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忽略略略

#1 Created at ... [Delete] [Delete and Lock User]

* coding: utf-8 *

import math

def quadratic(a,b,c): if a == 0: raise TypeError('a不能为0') if not isinstance(a,(int,float)) or not isinstance(b,(int,float)) or not isinstance(c,(int,float)): raise TypeError('Bad operand type') delta = math.pow(b,2) - 4ac if delta < 0: return '无实根' x1= (math.sqrt(delta)-b)/(2a) x2=-(math.sqrt(delta)+b)/(2a) return x1,x2 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')

忽略略略

#2 Created at ... [Delete] [Delete and Lock User]

import math

def quadratic(a,b,c): if not isinstance (a,(int,float)) or not isinstance (b,(int,float)) or not isinstance (c,(int,float)): raise TypeError ('error') if a == 0: raise TypeError ('a不等于0') d = b*2-4ac if d < 0: return '无解' x1=(math.sqrt(d)-b)/(2a) x2=-(math.sqrt(d)+b)/(2*a) return x1,x2 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')

南曦蔷

#3 Created at ... [Delete] [Delete and Lock User]

该怎么实现啊?


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