【2020蓝桥杯】python组真题解析 -凯发k8官方网

本资料整理者&代码编写者：夏2同学 个人邮箱：1754082565@qq.com 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][j1] == "0" \and l[i][j2] == "2" and l[i][j3] == "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[i1][j1] == "0" \and l[i2][j2] == "2" and l[i3][j3] == "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[i1][j] == "0" \and l[i2][j] == "2" and l[i3][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-d-d" % (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("d" % 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("d " % 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[j1]:# 交换lst[j], lst[j1] = lst[j1], 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 = n2nums = [[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, n1):line = map(int, input().split(" "))for j, e in enumerate(line):nums[i][j1] = 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, i1):# 如果下一行左边或者右边不为0if dp[i1][j] != 0 or dp[i1][j1] != 0:dp[i][j] = nums[i][j] max(dp[i1][j], dp[i1][j1])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(j1)))else:m[j] = p[1:].count(int(j1))'''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][2mm[k]]-r[k][1mm[k]]def down1(k):if mm[k] == 1:return r[k][2]elif mm[k] > r[k][1]:return 0else:return r[k][1mm[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][1mm[i]] print(point) '''得到总价值'''

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