【华为机试】2023年真题B卷（python）-乘坐保密电梯

一、题目

二、输入输出

三、示例

四、要求

五、参考代码 

# -*- coding: utf-8 -*-

params = [int(x) for x in input().split(" ")]
target = params[0]
count = params[1]
nums =  [int(x) for x in input().split(" ")]
#求和
sum_val = 0
min_distance = float("inf")
floor_list = []
for i in range(count):
    floor_list.append(nums[i])
    sum_val += nums[i]


def solve(target, sum_val, floor_list,min_distance, n, new_list, index,downs, flag):
    if(flag):
        return
    if(n != 0):
        for i in range(index, len(floor_list)):
            new_list.append(floor_list[i])
            solve(target,sum_val, floor_list,min_distance, n-1, new_list, i+1,downs,flag)
            new_list.pop(len(new_list) - 1)  
    else :
        cnt1 = 0
        for i in range(len(new_list)):
            cnt1 += new_list[i]
        
        distance = abs(sum_val - 2 * cnt1 - target)
        if(distance == 0):
            downs = new_list
            flag = true
        elif(min_distance > distance):
            min_distance = distance
            downs = new_list

#找出下降层数
downs = []
flag = False
solve(target, sum_val, floor_list,min_distance, count/2, [], 0,downs,flag)

for  down in downs :
    if(floor_list.index(down) != -1):
        floor_list.pop(floor_list.index(down))
    

floor_list.sort(reverse=True)
downs.sort(reverse=True)

result_str = ""
#上升和下降交替存在
for  i in range(len(floor_list)):
    result_str += str(floor_list[i]) + " "
    if(i < len(downs) ):
        result_str += str(downs[i]) + " "
    

print(result_str)