python 内部自带大整数运算能力，整数运算不会溢出，只要内存足够，就oK
下面的例子演示了两个32位整数加法的情况（通过位运算实现），为了模拟溢出的效果，必须人工的进行位运算，~运算符除了求反，还是二进制的补运算符，运算过后的二进制数字按照补码解释，例如 ~（0011 1100） = （1100 0011） = -61
def getSum(a, b):
"""
:type a: int
:type b: int
:rtype: int
"""
MAX = 0X7fffffff
MIN = 0X80000000
while b != 0 :
a,b = a^b,(a&b)<<1
print(" a = {0:b},b = {1:b}".format(a,b))
return a
def getSum_(a, b):
"""
:type a: int
:type b: int
:rtype: int
"""
MAX = 0x7FFFFFFF
MIN = 0x80000000
mask = 0xFFFFFFFF
while b != 0:
a, b = (a ^ b) & mask, ((a & b) << 1) & mask
print(type(a))
print(" a = {0:b},b = {1:b}".format(a,b))
return a if a <= MAX else ~(a^mask)
print(getSum_(-1,-1))
print(getSum(-1,1))
补充：python 循环内部添加多个条件判断会出现越界
1.循环遍历数组是,想添加条件修改时,只删除第一个
# -*- coding: utf-8 -*-
a=[11,22,33,44,55]
for i in a:
if i == 11 or i ==22:
a.remove(i)
for i in a:
print(i)
'''
33
55
[Finished in 0.1s]
'''
2.应该引入被删除为一个数组
# -*- coding: utf-8 -*-
a=[11,22,33,44,55]
b=[]
for i in a:
if i == 11 or i ==22:
b.append(i)
for i in b:
a.remove(i)
for i in a:
print(i)
'''
33
44
55
[Finished in 0.1s]
'''
