蓝桥python考题_蓝桥杯python2021?

问题描述

思路
核心：寻找每一号位的最大值

枚举，暴力破解缩小范围，寻找每一号位5个最大评分，再进行for循环

代码如下

import os
import sys

# 请在此输入您的代码
s1=[97,92,0,0,89,82,0,0,0,95,0,0,94,0,0,0,98,93,0,0]
s2=[90,85,0,0,83,86,0,97,0,99,0,0,91,83,0,0,83,87,0,99]
s3=[0,96,0,0,97,0,0,96,89,0,96,0,0,87,98,0,99,92,0,96]
s4=[0,0,0,80,0,0,87,0,0,0,97,93,0,0,97,93,98,96,89,95]
s5=[0,0,93,86,0,0,90,0,0,0,0,98,0,0,98,86,81,98,92,81]

ls=[]
'''
#暴力求解:
for i in s1:
    for j in s2:
        for k in s3:
            for x in s4:
                for y in s5:
                    if(i!=j!=k!=x!=y):
                        ls.append(i+j+k+x+y)

print(max(ls))
'''

#范围求解:
s1_5=[]
s2_5=[]
s3_5=[]
s4_5=[]
s5_5=[]
#分别存入n号位对应的5个最大数值
for i in range(0,5):
    s1_5.append(max(s1))
    s1.remove(max(s1))
for i in range(0,5):
    s2_5.append(max(s2))
    s2.remove(max(s2))
for i in range(0,5):
    s3_5.append(max(s3))
    s3.remove(max(s3))
for i in range(0,5):
    s4_5.append(max(s4))
    s4.remove(max(s4))
for i in range(0,5):
    s5_5.append(max(s5))
    s5.remove(max(s5))


for i in s1_5:
    for j in s2_5:
        for k in s3_5:
            for x in s4_5:
                for y in s5_5:
                    if(i!=j!=k!=x!=y):
                        ls.append(i+j+k+x+y)
print(ls)
print(max(ls))

import os
import sys

# 请在此输入您的代码
# a,b,c,d = 1,1,1,0

# for i in range(3,20190324):
#     d = (a+b+c)%10000
#     a,b,c = b,c,d

# print(d)

print(4659)

import os
import sys
import math
# ①构造列表，存储质数
# ②质数个数等于2019，放回其值
# n = sys.maxsize
# ls = [2,3]
# for i in range(4,n):
#     for j in range(2,i//2+1):
#         if i%j == 0:
#             break
#     else:ls.append(i)
#     if len(ls) == 2019:
#         break
# print(ls[2018])
print(17569)

import os
import sys
# 个位
a = 2019 % 26
b = 2019 // 26 
# 十位
c = b % 26
# 百位
d = b // 26
list = '*ABCDEFGHIJKLMNOPQRSTUVWXYZ'
print(list[d] + list[c] + list[a])

import os
import sys
# 要求：分成不同的正整数，不包含2，4，不重复性
# 1、利用for循环，
# 2、不包含2，4 利用str(i).count(2) == 0
ans = 0
for i in range(1,672):
  for j in range(i+1,(2019-i)//2+1):
    a = str(i)+str(j)+str(2019-i-j)
    if 2019 - i - j > j and '2' not in a and '4' not in a:
      ans += 1 
print(ans)

问题描述

个人思路

先判断是2019的整倍数将数，转换为str型，对其for循环遍历，除于2若各位数都为奇数，则输出，并跳出循环

代码

import os
import sys
n = sys.maxsize
for i in range(n):
  if i % 2019 ==0:
    for j in str(i):
      if int(j) % 2 == 0:
        break
    else:
      print(i)
      break

问题描述

代码如下

import os
import sys
import math

N = int(input())
A = list(map(int, input().split()))
ans = []
#ceil向上取整，log求对数
deep = math.ceil(math.log(N+1,2))
b = 0
for i in range(1, deep+1):
    c = 2 ** i - 1
    ans.append(sum(A[b:c]))
    b = c

#index查找索引值，默认0开始
print(ans.index(max(ans))+1)

参考

注：临时变量存储总和的时间 < 列表存储数据

【填空】
1、枚举爆破，for循环
2、找规律，容易数据溢出，缩小范围
3、枚举暴力
4、取余，获取每个数的每个位数
5、
【程序设计】
1、数学逻辑知识