代码中的# -*- coding: utf-8 -*-和print u'''...'''中的u造成重复,导致结果输出不正常。这个题的坑挖的好 ~-~ !
2016-08-04
最新回答 / 815638320
L = []for m in 'ABC': for n in '123': for o in 'ABC': L.append(m+n+o)if m == o: print L
2016-08-03
最新回答 / icediv
这种式子分3部分,第一部分是 x 第二部分是 for后面的,第三部分是 if后面的str方法将x变成字符串处理,由于for后面的区间是100到1000的三位数数字。str(x)[0] 取首位数字的字符内容,str(x)[-1]取最后一个位置的字符内容,这个过滤条件就可以得到对称数字的list了
2016-08-03
def quadratic_equation(a, b, c):
x = b*b-4*a*c
if x < 0:
return none
elif x == 0:
return -b / (2 *a)
else:
y1 = -b + math.sqrt(x)
y2 = -b - math.sqrt(x)
z1 = y1 / (2*a)
z2 = y2 / (2*a)
return z1, z2
x = b*b-4*a*c
if x < 0:
return none
elif x == 0:
return -b / (2 *a)
else:
y1 = -b + math.sqrt(x)
y2 = -b - math.sqrt(x)
z1 = y1 / (2*a)
z2 = y2 / (2*a)
return z1, z2
2016-08-03
def square_of_sum(L):
sum = 0
for x in L:
sum += x**2
return sum
print square_of_sum([1, 2, 3, 4, 5])
print square_of_sum([-5, 0, 5, 15, 25])
sum = 0
for x in L:
sum += x**2
return sum
print square_of_sum([1, 2, 3, 4, 5])
print square_of_sum([-5, 0, 5, 15, 25])
2016-08-03
L = []
x = 1
while x <= 100:
L.append(x**2)
x += 1
print sum(L)
x = 1
while x <= 100:
L.append(x**2)
x += 1
print sum(L)
2016-08-03
s = set(['Adam', 'Lisa', 'Paul'])
L = ['Adam', 'Lisa', 'Bart', 'Paul']
for name in L:
if name in s:
s.remove(name)
else:
s.add(name)
print s
L = ['Adam', 'Lisa', 'Bart', 'Paul']
for name in L:
if name in s:
s.remove(name)
else:
s.add(name)
print s
2016-08-03
s = set(['adam','bart'])
print 'adam' in s
print 'bart' in s
print 'adam' in s
print 'bart' in s
2016-08-03