【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
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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：624total = 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：16520def search_row(l):res = 0m = 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 += 1return resdef search_xie(l):res = 0m = 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 += 1return resdef search_col(l):res = 0m = 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 reswith open("2020.txt",'r') as fp:lines = fp.readlines()i = 0for line in lines:lines[i] = line.strip('\n')i+=1# 16520print(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：8879flag = 6def nextDay(date):'''date: 格式 yyyy-mm-ddreturn next_date,is_monday,is_month_begin星期一和月初'''date = list(map(int,date.split("-")))y,m,d = dated+=1if m == 12 and d == 32:y += 1m = 1d = 1elif m == 2:if is_leap_year(y):if d == 30:m = 3d = 1else:if d == 29:m = 3d = 1elif is_big_month(m):if d == 32:m += 1d = 1elif not is_big_month(m):if d == 31:m += 1d = 1global flagadd_week()is_month_begin = True if d == 1 else Falseis_monday = True if flag == 1 else Falsenew_date = "%d-%02d-%02d" % (y,m,d)return new_date,is_monday,is_month_begindef is_leap_year(year):if year % 4 == 0 and year % 100 != 0:return Trueelif year % 400 == 0:return Truereturn Falsedef is_big_month(month):if month in [1,3,5,7,8,10,12]:return Trueelse:return Falsedef add_week():global flagflag = (flag + 1) % 8 if (flag + 1) % 8 else 1def main():start_time = "2000-01-01"# end_time = "2000-01-10"end_time = "2020-10-01"total = 2while start_time != end_time:start_time,is_monday,is_month_begin = nextDay(start_time)if is_monday or is_month_begin:total += 2else:total += 1print(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：761def 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 = 1res = [[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 += 1res[i][j] = numwhile j > 0:# 向斜左下角num += 1i += 1j -= 1res[i][j] = numnum += 1# 向下i += 1res[i][j] = numwhile i > 0:# 向斜右上角num += 1i -= 1j += 1res[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：jonmlkihgfedcbaimport randomcount = 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 += 1def bubble_sort(lst):n = len(lst)for i in range(n-1):is_ordered = Truej = 0while j < n-i-1:if lst[j] > lst[j+1]:# 交换lst[j], lst[j+1] = lst[j+1], lst[j]global countcount += 1is_ordered = Falsej += 1if 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/learningfrom decimal import *n = int(input())socres = [int(input()) for i in range(n)]youxiu = 0
jige = 0for i in socres:if i >= 85:youxiu += 1if 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] * 26string = input()for char in string:charLst[ord(char)-97] += 1max_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 = 105n = int(input())MAX = n+2nums = [[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0if 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 syss = []
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 0elif 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 0else: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:returnwhile max(up[:n]) > down[n] and mm[n] > 0:t = up.index(max(up[:n]))mm[t] += 1up[t] = up1(t)mm[n] -= 1down[n] = down1(n)main1(n-1)returnpoint = 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'''

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