刷题记录

python 的字符串数学表达式计算题，如果时间不够使用：

eval(str)

有一个长为n的数组A，求满足0≤a≤b

# -*- coding:utf-8 -*-
class LongestDistance:
    def getDis(self, A, n):
        # write code here
        maxdiff = 0
        minValue = A[0]
        for i in xrange(n):
            minValue = min(minValue,A[i])
            maxdiff = max(maxdiff,A[i]-minValue)
        return maxdiff

输入一个包含6个数字的数据，输出24时间进制的数值：

list1 = ['0','2','3','0','5','6']
all_index = []
all_num =  []
exclude = []
time = []
max1 = 0
max2 = 0
max3 = 0
tmp = [0,0]
a = 0
for i in range(6):
    for j in range(i+1,6):
        all_index.extend([[i,j],[j,i]])
print(all_index)
for index in all_index:
        num = list1[index[0]] + list1[index[1]]
        num = int(num)
        if num >= max1 and num <= 24:
            max1 = num
            tmp[0]=index[0]
            tmp[1]=index[1]
            a +=1
exclude.extend(tmp)
for index in all_index:
    num = list1[index[0]] + list1[index[1]]
    num = int(num)
    print(num)
    if num >= max2 and num <= 59 and (index[0] not in exclude) and (index[1] not in exclude):
        max2 = num
        tmp[0]=index[0]
        tmp[1]=index[1]
        a +=1
exclude.extend(tmp)

for index in all_index:
    num = list1[index[0]] + list1[index[1]]
    num = int(num)
    if num >= max3 and num <= 60 and (index[0] not in exclude) and (index[1] not in exclude):
        max3 = num
        a +=1

if a>2:
    print(max1, max2, max3)
    print(exclude)
else:
    print('invalid')

牛客python3 大佬的刷题记录：
华科平凡

华为笔试题

有一个数组a[N]顺序存放0~N-1，要求每隔两个数删掉一个数，到末尾时循环至开头继续进行，求最后一个被删掉的数的原始下标位置。以8个数(N=7)为例:｛0，1，2，3，4，5，6，7｝，0->1->2(删除)->3->4->5(删除)->6->7->0(删除),如此循环直到最后一个数被删除。

输入描述:

每组数据为一行一个整数n(小于等于1000)，为数组成员数,如果大于1000，则对a[999]进行计算。

输出描述:

一行输出最后一个被删掉的数的原始下标位置。

示例1

输入

8

输出

6

约瑟夫环解法

while True:
    try:
        n = int(input())
    except:
        exit()
    a = list(range(n))
    i = 0
    while len(a) > 1:
        i = (i+2) % len(a)
        if i != len(a) - 1:
            a = a[:i] + a[i+1:]
        else:
            a = a[:i]
            i = 0
    print (a[0])