【2020蓝桥杯】Python组真题解析 - 第十一届蓝桥杯
本资料整理者&代码编写者：夏2同学
个人邮箱：[email protected]
请勿未经同意转载 & 如有错误，欢迎指正
资料下载：https://download.csdn.net/download/xia_yanbing/16742326
备注：目前蓝桥杯Python编辑器是：IDLE。
如果您想了解如何使用，可以参考这篇。
https://blog.csdn.net/xia_yanbing/article/details/114641646
目录
-
- A-门牌制作.py
-
- 参考答案：624
- B-寻找2020.py
-
- 参考答案：16520
- C-跑步锻炼.py
-
- 参考答案：8879
- D-蛇形填数.py
-
- 参考答案：761
- E-排序.py
-
- 参考答案：jonmlkihgfedcba
- F-成绩统计.py
- G-单词分析.py
- H-数字三角形.py
- I-平面分割.py
- J-装饰珠.py
A-门牌制作.py
题目地址：https://www.lanqiao.cn/problems/592/learning/
参考答案：624
# https://www.lanqiao.cn/problems/592/learning/
# 参考答案：624
total = 0
for i in range(1,2021):
total += (str(i).count("2"))
print(total)
# 624
B-寻找2020.py
题目地址：无在线练习题地址
参考答案：16520
# 无在线练习题地址
# 参考答案：16520
def search_row(l):
res = 0
m = len(l[0])
n = len(l)
for i in range(n):
for j in range(m-3):
if l[i][j] == "2" and l[i][j+1] == "0" \
and l[i][j+2] == "2" and l[i][j+3] == "0":
res += 1
return res
def search_xie(l):
res = 0
m = len(l[0])
n = len(l)
for i in range(n-3):
for j in range(m-3):
if l[i][j] == "2" and l[i+1][j+1] == "0" \
and l[i+2][j+2] == "2" and l[i+3][j+3] == "0":
res += 1
return res
def search_col(l):
res = 0
m = len(l[0])
n = len(l)
for j in range(m):
for i in range(n-3):
if l[i][j] == "2" and l[i+1][j] == "0" \
and l[i+2][j] == "2" and l[i+3][j] == "0":
res += 1
return res
with open("2020.txt",'r') as fp:
lines = fp.readlines()
i = 0
for line in lines:
lines[i] = line.strip('\n')
i+=1
# 16520
print(search_row(lines)+search_xie(lines)+search_col(lines))
C-跑步锻炼.py
题目地址：https://www.lanqiao.cn/problems/597/learning/
参考答案：8879
# https://www.lanqiao.cn/problems/597/learning/
# 参考答案：8879
flag = 6
def nextDay(date):
'''
date: 格式 yyyy-mm-dd
return next_date,is_monday,is_month_begin
星期一和月初
'''
date = list(map(int,date.split("-")))
y,m,d = date
d+=1
if m == 12 and d == 32:
y += 1
m = 1
d = 1
elif m == 2:
if is_leap_year(y):
if d == 30:
m = 3
d = 1
else:
if d == 29:
m = 3
d = 1
elif is_big_month(m):
if d == 32:
m += 1
d = 1
elif not is_big_month(m):
if d == 31:
m += 1
d = 1
global flag
add_week()
is_month_begin = True if d == 1 else False
is_monday = True if flag == 1 else False
new_date = "%d-%02d-%02d" % (y,m,d)
return new_date,is_monday,is_month_begin
def is_leap_year(year):
if year % 4 == 0 and year % 100 != 0:
return True
elif year % 400 == 0:
return True
return False
def is_big_month(month):
if month in [1,3,5,7,8,10,12]:
return True
else:
return False
def add_week():
global flag
flag = (flag + 1) % 8 if (flag + 1) % 8 else 1
def main():
start_time = "2000-01-01"
# end_time = "2000-01-10"
end_time = "2020-10-01"
total = 2
while start_time != end_time:
start_time,is_monday,is_month_begin = nextDay(start_time)
if is_monday or is_month_begin:
total += 2
else:
total += 1
print(start_time,is_monday,is_month_begin,total)
print(total)
# ans =
main()
D-蛇形填数.py
题目地址：https://www.lanqiao.cn/problems/594/learning/
参考答案：761
# https://www.lanqiao.cn/problems/594/learning/
# 参考答案：761
def print_out(lst):
# 一行行输出
for line in lst:
for x in line:
print("%04d" % x,end=" ")
# print(" ".join(map(str, i)))
print()
lst = [i for i in range(1000)]
num = 1
res = [[0 for i in range(101)] for i in range(101)]
i, j = 0, 0
#
res[0][0] = num
while i != 100 and j != 100:
num += 1
# 向右
j += 1
res[i][j] = num
while j > 0:
# 向斜左下角
num += 1
i += 1
j -= 1
res[i][j] = num
num += 1
# 向下
i += 1
res[i][j] = num
while i > 0:
# 向斜右上角
num += 1
i -= 1
j += 1
res[i][j] = num
# 蛇形填数字
with open ('./out.txt','w+') as fp:
for line in res:
for x in line:
fp.write("%04d " % x)
fp.write("\n")```
E-排序.py
题目地址：https://www.lanqiao.cn/problems/598/learning/
参考答案：jonmlkihgfedcba
# https://www.lanqiao.cn/problems/598/learning/
# 参考答案：jonmlkihgfedcba
import random
count = 0
# def bubble_sort(lst):
# n = len(lst)
# for i in range(n-1):
# # is_ordered = True
# j = 0
# while j < n-i-1:
# if lst[j] > lst[j+1]:
# # 交换
# lst[j], lst[j+1] = lst[j+1], lst[j]
# global count
# count += 1
# # is_ordered = False
# j += 1
def bubble_sort(lst):
n = len(lst)
for i in range(n-1):
is_ordered = True
j = 0
while j < n-i-1:
if lst[j] > lst[j+1]:
# 交换
lst[j], lst[j+1] = lst[j+1], lst[j]
global count
count += 1
is_ordered = False
j += 1
if is_ordered:
# 结束冒泡
break
# string =[chr(i) for i in range(ord('a'),ord('p'))]
# data = ['o', 'n', 'm', 'l', 'k', 'j',
# 'i', 'h', 'g', 'f', 'e', 'd', 'c', 'b', 'a']
# 1. 最短 2. 字典序低
data = ['j', 'o', 'n', 'm', 'l', 'k',
'i', 'h', 'g', 'f', 'e', 'd', 'c', 'b', 'a']
# data = string[::-1]
print(data)
bubble_sort(data)
print(data)
print(count)
F-成绩统计.py
题目地址：https://www.lanqiao.cn/problems/502/learning
# https://www.lanqiao.cn/problems/502/learning
from decimal import *
n = int(input())
socres = [int(input()) for i in range(n)]
youxiu = 0
jige = 0
for i in socres:
if i >= 85:
youxiu += 1
if i >= 60:
jige += 1
# 高精度
percent1 = int(round(float(Decimal(jige) / Decimal(n)) * 100, 0))
percent2 = int(round(float(Decimal(youxiu) / Decimal(n)) * 100, 0))
print("%d%%" % percent1)
print("%d%%" % percent2)
G-单词分析.py
题目地址：https://www.lanqiao.cn/problems/504/learning
# https://www.lanqiao.cn/problems/504/learning
# 记录26个英文字母的顺序
charLst = [0] * 26
string = input()
for char in string:
charLst[ord(char)-97] += 1
max_value = max(charLst)
res_index = charLst.index(max_value)
res_chr = chr(res_index + 97)
print(res_chr)
print(max_value)
H-数字三角形.py
题目地址：https://www.lanqiao.cn/problems/505/learning/
# https://www.lanqiao.cn/problems/505/learning/
# 借鉴了 https://www.jianshu.com/p/c20b6b9a178a （C++）
if __name__ == '__main__':
# MAX = 105
n = int(input())
MAX = n+2
nums = [[0 for i in range(1, MAX)] for i in range(1, MAX)]
dp = [[0 for i in range(1, MAX)] for i in range(1, MAX)]
# 读取用户输入
for i in range(1, n+1):
line = map(int, input().split(" "))
for j, e in enumerate(line):
nums[i][j+1] = e
# print(nums)
# 初始化最后一行
if n % 2 == 0:
dp[n][n//2] = nums[n][n//2]
dp[n][n//2 + 1] = nums[n][n//2 + 1]
else:
dp[n][n//2 + 1] = nums[n][n//2 + 1]
# 从后往前依次遍历
for i in range(n-1, 0, -1):
for j in range(1, i+1):
# 如果下一行左边或者右边不为0
if dp[i+1][j] != 0 or dp[i+1][j+1] != 0:
dp[i][j] = nums[i][j] + max(dp[i+1][j], dp[i+1][j+1])
print(dp[1][1])
I-平面分割.py
题目地址：https://www.lanqiao.cn/problems/503/learning/
# https://www.lanqiao.cn/problems/503/learning/
# 思路：画几张图观察一下可以发现一下特点：
# 1、重边不会影响区域数目。
# 2、每新加入一条边只要不是重边区域数目必定+1。
# 3、加入的新边如果与之前加入的边每有一个交点则区域数目+1。
n = int(input())
# 存放直线数（必须得不重合）
set_line = set()
# 存放交点数
set_point = set()
for i in range(n):
tmp = tuple(input().split())
set_line.add(tmp)
set_point.add(tmp[0])
print(len(set_line)+len(set_point))
J-装饰珠.py
题目地址：https://www.lanqiao.cn/problems/507/learning/
# https://www.lanqiao.cn/problems/507/learning/
import os
import sys
s = []
r = []
m = []
for i in range(6):
s.append(list(map(int, input().split())))
n = int(input())
for i in range(n):
r.append(list(map(int, input().split())))
for i in range(6):
p = s[i]
for j in range(n):
if i == 0:
m.append(p[1:].count(int(j+1)))
else:
m[j] += p[1:].count(int(j+1))
'''print(m)''' '''用m[j]记录6件装备共有 j+1级孔m[j]个'''
mm = m
'''假设j级孔全用j级珠子'''
up = []
down = []
'''up记录加一颗珠子提升的价值，down记录少一颗珠子减少的价值'''
def up1(k):
if mm[k] >= r[k][1]:
return 0
elif mm[k] == 0:
return r[k][2]
else:
return r[k][2+mm[k]]-r[k][1+mm[k]]
def down1(k):
if mm[k] == 1:
return r[k][2]
elif mm[k] > r[k][1]:
return 0
else:
return r[k][1+mm[k]]-r[k][mm[k]]
for i in range(n):
up.append(up1(i))
down.append(down1(i))
'''print(up,down)'''
'''第n+1级孔，如果1到n级孔中珠子加1的价值比它减少1颗的价值大，则数值各加1、减1，up、down更新'''
def main1(n):
if n == 0:
return
while max(up[:n]) > down[n] and mm[n] > 0:
t = up.index(max(up[:n]))
mm[t] += 1
up[t] = up1(t)
mm[n] -= 1
down[n] = down1(n)
main1(n-1)
return
point = 0
main1(n-1)
'''输入有n级，为了和形参统一，此处减1'''
'''print(up,down)
print(mm)'''
for i in range(n):
if mm[i] != 0:
point += r[i][1+mm[i]]
print(point)
'''得到总价值'''