x1 = 1
d = 3
n = 100
x100 = x1+d*(n-1)
s = (x1+x100)*n/2
print s
d = 3
n = 100
x100 = x1+d*(n-1)
s = (x1+x100)*n/2
print s
2016-05-29
def firstCharUpper(s):
return s.upper()[:1]+s[-(len(s)-1):]
print firstCharUpper('hello')
print firstCharUpper('sunday')
print firstCharUpper('september')
return s.upper()[:1]+s[-(len(s)-1):]
print firstCharUpper('hello')
print firstCharUpper('sunday')
print firstCharUpper('september')
2016-05-29
x1 = (-b + math.sqrt(delta)) / (2.0 * a), (2.0 * a)你写成除以(2 * a)就错了
2016-05-29
题目中t = ('a', 'b', ['A', 'B'])元组是包含三个元素的,如果改为t = ('a', 'b', 'A', 'B')则元组变为四个元素了。肯定报错啦
2016-05-29
sum = 0
x = 1
while x < 100:
sum = sum + x + 2
print sum
x = 1
while x < 100:
sum = sum + x + 2
print sum
2016-05-29
用切片操作符就一句
L = ['Adam','lisa','Bart']
L = L[::-1]
print L
L = ['Adam','lisa','Bart']
L = L[::-1]
print L
2016-05-29
转义字符表示该字符是一个普通字符,不是字符串的起始,'Bob said \"I\'m OK\".'
2016-05-28
自己写得,完全靠脑子,有点繁琐,不过还行吧,哈哈
def square_of_sum(L):
I = []
for a in L:
a = a * a
I.append(a)#把L列表的各项平方添加到I列表
return sum(I)#得出I列表的总和,打印
print square_of_sum([1, 2, 3, 4, 5])
print square_of_sum([-5, 0, 5, 15, 25])
def square_of_sum(L):
I = []
for a in L:
a = a * a
I.append(a)#把L列表的各项平方添加到I列表
return sum(I)#得出I列表的总和,打印
print square_of_sum([1, 2, 3, 4, 5])
print square_of_sum([-5, 0, 5, 15, 25])
2016-05-28