我要做的是写一个二次方程求解器但是当解决方案应该是-1
时,就像在quadratic(2, 4, 2)
中一样,它返回1
我做错了什么?
#!/usr/bin/python import math def quadratic(a, b, c): #a = raw_input("What\'s your `a` value?\t") #b = raw_input("What\'s your `b` value?\t") #c = raw_input("What\'s your `c` value?\t") a, b, c = float(a), float(b), float(c) disc = (b*b)-(4*a*c) print "Discriminant is:\n" + str(disc) if disc >= 0: root = math.sqrt(disc) top1 = b + root top2 = b - root sol1 = top1/(2*a) sol2 = top2/(2*a) if sol1 != sol2: print "Solution 1:\n" + str(sol1) + "\nSolution 2:\n" + str(sol2) if sol1 == sol2: print "One solution:\n" + str(sol1) else: print "No solution!"
编辑:它返回以下内容......
>>> import mathmodules >>> mathmodules.quadratic(2, 4, 2) Discriminant is: 0.0 One solution: 1.0
答案 0 :(得分:12)
除非我上学后公式发生了变化(人们永远不能确定),所以(-b +- sqrt(b^2-4ac)) / 2a
,你的代码中有b
。
[编辑]我可以建议一个重构吗?
def quadratic(a, b, c):
discriminant = b**2 - 4*a*c
if discriminant < 0:
return []
elif discriminant == 0:
return [-b / (2*a)]
else:
root = math.sqrt(discriminant)
return [(-b + root) / (2*a), (-b - root) / (2*a)]
print quadratic(2, 3, 2) # []
print quadratic(2, 4, 2) # [-1]
print quadratic(2, 5, 2) # [-0.5, -2.0]
答案 1 :(得分:2)
二次方的解是
x = (-b +/- sqrt(b^2 - 4ac))/2a
但你编码的是
x = (b +/- sqrt(b^2 - 4ac))/2a
这就是你收到签名错误的原因。
答案 2 :(得分:1)
top1
和top2
的迹象有误,请参阅http://en.wikipedia.org/wiki/Quadratic_equation
答案 3 :(得分:0)
top1 = b + root
top2 = b - root
应该是:
top1 = -b + root
top2 = -b - root