import math
def quadratic_equation(a, b, c):
d = b*b - 4*a*c
if d > 0:
return (-b+math.sqrt(d))/(2*a), (-b-math.sqrt(d))/(2*a)
elif d == 0:
return (-b)/(2*a)
else:
return '无解'
print quadratic_equation(2, 3, 0)
print quadratic_equation(1, -6, 5)
def quadratic_equation(a, b, c):
d = b*b - 4*a*c
if d > 0:
return (-b+math.sqrt(d))/(2*a), (-b-math.sqrt(d))/(2*a)
elif d == 0:
return (-b)/(2*a)
else:
return '无解'
print quadratic_equation(2, 3, 0)
print quadratic_equation(1, -6, 5)
2016-08-24
def square_of_sum(L):
sum=0
for num in L:
sum=sum+num*num
return sum
print square_of_sum([1, 2, 3, 4, 5])
print square_of_sum([-5, 0, 5, 15, 25])
sum=0
for num in L:
sum=sum+num*num
return sum
print square_of_sum([1, 2, 3, 4, 5])
print square_of_sum([-5, 0, 5, 15, 25])
2016-08-24
# 如果要加入很多元素,建议使用以下代码
s = set(['Adam', 'Lisa', 'Paul'])
L = ['Adam', 'Lisa', 'Bart', 'Paul']
for x in s:
for y in L:
if x ==y:
pass
else:
s.add(y)
print s
s = set(['Adam', 'Lisa', 'Paul'])
L = ['Adam', 'Lisa', 'Bart', 'Paul']
for x in s:
for y in L:
if x ==y:
pass
else:
s.add(y)
print s
2016-08-24
s = set(['adam','lisa','bart','paul'])
print 'adam' in s
print 'bart' in s
这样就可以了啊,为什么要写那么复杂
print 'adam' in s
print 'bart' in s
这样就可以了啊,为什么要写那么复杂
2016-08-24
d = {
'Adam': 95,
'Lisa': 85,
'Bart': 59
}
for k in d:
print '%s:'%k,d[k]
'Adam': 95,
'Lisa': 85,
'Bart': 59
}
for k in d:
print '%s:'%k,d[k]
2016-08-24
真是的 我想应该搞懂了一些了
评判系统首先是要结果对 还要出现上面定义的才算通关.................
评判系统首先是要结果对 还要出现上面定义的才算通关.................
2016-08-24
d = {
'Adam':95,
'Lisa':85,
'Paul':75,
'Bart':59
}
这样也是对的吧,但是确实最后不加标点,提交就是错误的!
'Adam':95,
'Lisa':85,
'Paul':75,
'Bart':59
}
这样也是对的吧,但是确实最后不加标点,提交就是错误的!
2016-08-24