"""第一题"""
def f(n):
res = 0
for i in range(1, n+1, 2):
res += i
for i in range(2, n+1, 2):
res -= i
return res
"""第二题"""
res = []
nums = ['9','8','7','6','5','4','3','2','1']
def convert(formula):
if not formula or len(formula) == 0:
return 0
tmp = ''
for i in range(len(formula)):
if formula[i] == '+' formula[i] != '+' and formula[i] != '-':
return int(tmp) + valid(formula[i+1:])
elif formula[i] == '-':
return int(tmp) - valid(formula[i+1:])
else:
tmp += formula[i]
def valid(formula):
if convert(formula) == 100:
return True
return False
def cal(formu, idx):
if idx == len(nums) - 1:
formula = ''.join(formu)
if valid(formula):
res.append(formula)
for i in range(idx, len(nums)-1):
cal(formu, i+1)
cal(formu[:i] + [formu[i]+'+'] + formu[i+1:], i+1)
cal(formu[:i] + [formu[i]+'-'] + formu[i+1:], i+1)
cal(nums, 0)
print('Results of formulas below all equal to 100')
for i in res:
print(i)
"""第三题"""
假设我们开始排列是1,2,3,4,5
然后首先我们要让第一个位置变成2，3，4或者5才满足条件，而在其中我们又会遇到原位置没有变化的情况，则我们知道每次变换2个位置的时候总有重复，因为两边计算出的所有情况中总有一半是不满足的，并且是对于所有位置不同的情况下，那么我们不妨从第一个位置开始，则公式为 ，其中n为小朋友的个数
开始编码：
from math import factorial
def cal(n):
return (n-1) * factorial(n-1) // 2
"""第四题"""
import sys
def maxSum(nums, window_len):
res = -sys.maxsize
for i in range(len(nums)-window_len):
res = max(res, sum(nums[i:i+k]))
return res
"""第五题"""
class ListNode(object):
def __init__(self, val, next, sib):
self.val = val
self.next = next
self.sib = sib
class Clone(object)
def cloneComplicateListNode(self, head):
if not head:
return head
dummy = cur = ListNode(head.val, None, None)
while head:
cur.next = self.cloneComplicateListNode(head.next)
cur.sib = self.cloneComplicateListNode(head.sib)
head = head.next
cur = cur.next
return dummy
