商汤笔试20190819

#!/usr/bin/env python
# coding: utf-8

code1:
# 解析逆波兰式子
# 通过100%
def evalRPN(tokens):
    t,f = tokens.pop(),evalRPN
    if t in ‘+-*/‘:
        b,a = f(tokens),f(tokens)
        t = eval("a" + t + "b")
    return int(t)

if __name__ == "__main__":
    tokens = input()
    tokens = list([n for n in tokens.split()])
#     tokens = ["2","1","+","3","*"]
    print(evalRPN(tokens))

code2:
# 最高的山峰
# 通过78%
def longestMountain(N,A):
    res = 0
    i = 0
    while i <N:
        j = i
        #找到上升的终点，即山顶
        while j+1<len(A) and A[j]<A[j+1]:
            j += 1
        mid = j
        #找到下降的终点
        while j+1<len(A) and A[j]>A[j+1]:
            j += 1
        #判断是否构成山脉
        if i <mid<j:
            res = max(res,j-i+1)
        #若为平山，i后移，否则下降的终点作为起点
        if i == j:
            i += 1
        else:
            i = j
    return res        

if __name__ == "__main__":
    N = int(input())
    A = input()
    A = list([int(n) for n in A.split()])
#     N = 7
#     A = [9,2,5,8,4,1,7]
    print(longestMountain(N,A))

code3
# 谈学姐与投资组合
# f(i) = 0, i<0
# f(i) = f(0), i=0
# f(i) = a*f(i-1)+b*f(i-2)+c*f(i-3)+2*i**2-i+32767
# 输入n,a,b,c,f(0)
# 输出：f(n)
# 类似斐波那契数列，只通过了10%
def moneySum( n,a,b,c,f0):
    if n<0:
        return 0
    if n == 0:
        return f0
    if n == 1:
        f1 = a*f0+32767+2-2
        return f1
    if n == 2:
        f1 = a*f0+32767+2-2
        f2 = a*f1+b*f0+32767+2*n**2-n
        return f2
    return  a*moneySum(n-1,a,b,c,f0)+b*moneySum(n-2,a,b,c,f0)+c*moneySum(n-3,a,b,c,f0)+2*n**2-n+32767   

if __name__ == "__main__":
    array = input()
    array = list([int(n) for n in array.split()])
    n,a,b,c,f0 = array[0],array[1],array[2],array[3],array[4]
#     n,a,b,c,f0 = [10,0,0,0,100]
    print(moneySum(n,a,b,c,f0))

code4
#完全平方数组的数目：类似LeetCode_996_NumberOfSquarefulArrays
#https://blog.csdn.net/qq_17550379/article/details/87687106
#下列代码通过了87%
def numSquarefulPerms(N,A):
    nums = A
    nums.sort()
    res = []
    def backtrack(nums,tmp):
        if not nums:
            res.append(tmp)
            return None
#         print("len(nums)",len(nums))
        for i in range(len(nums)):
            #去重
            if i and nums[i] == nums[i-1]:
                continue
            #剪枝
            if not tmp or ((tmp[-1]+nums[i])**0.5)%1 == 0:
                backtrack(nums[:i]+nums[i+1:],tmp+[nums[i]])
    backtrack(nums,[])
    return len(res)            

if __name__ == "__main__":
#     N = int(input())
#     A = input()
#     A = list([int(n) for n in A.split()])
    N = 3
    A = [3,10,6]
    print(numSquarefulPerms(N,A))

原文地址：https://www.cnblogs.com/ivyharding/p/11380070.html