1. 门牌制作

1. 思路：从1到2020，查看之间有多少个2即可
2. 代码实现：

ls = []
for i in range(1, 2021):ls.append(str(i))
ans = 0
for i in ls:ans += i.count('2')
print(ans)

3. 答案为：624

2. 既约分数

1. 思路：暴力遍历即可
2. 代码实现：

import math
ans = 0
for i in range(1, 2021):for j in range(1, 2021):if math.gcd(i, j) == 1:ans += 1
print(ans)

3. 答案：2481215

3. 蛇形填数

1. 思路：找规律，根据处在行列相同的数字【1， 5， 13， 25…】可以看出，中间每个数相差等差数列个4，直接按等差数列求和，找到第一个数到第20行20列的数字相差多少个4即可
2. 代码实现：

count = 0
for i in range(1, 20):count += i
ans = 1 + count * 4
print(ans)

3. 答案：761

4. 跑步锻炼

1. 思路：找到2000.1.1到2020.10.1一共有多少天，其中有多少是周一或者月初，有多少既是周一又是月初即可。是由datetime模块即可
2. 代码实现：

import datetime
start_time = datetime.date(2000, 1, 1)
end_time = datetime.date(2020, 10, 1)
# 得到两者之间相隔的天数    7579
DeltaTime = end_time - start_time
aday = datetime.timedelta(days = 1)
ans = 0
for i in range(7579):if start_time.day == 1 or start_time.isoweekday() == 1:ans += 2else:ans += 1start_time += aday
# 最后一天没有计算，所以手动加2即可
ans += 2
print(ans)

3. 答案:8879

5. 七段码

1. 思路;从图的角度入手，将紧邻的边看作图中的节点的连通，美剧可能出现的情况，判断连通性即可
2. 代码实现：

import itertoolsnew_list = []
# 计数
num = 0
# 采用字典将该标号相接的所有标号列举出来
dict = {'a': ['f', 'b'], 'b': ['a', 'c', 'g'], 'c': ['b', 'd', 'g'],'d': ['e', 'c'], 'e': ['d', 'f', 'g'], 'f': ['a', 'e', 'g'],'g': ['b', 'c', 'e', 'f']}string = "".join(list(dict.keys()))
for i in range(1, 8):# 对string字符串中的前i个字符进行无序组合for j in (itertools.combinations(list(string), i)):new_list.append("".join(j))
for str1 in new_list:if len(str1) == 1:num += 1continuefor situation in itertools.permutations(str1):for c in range(1, len(situation)):if situation[c - 1] not in dict[situation[c]]:breakelse:num += 1break
print(num)

4. 答案：80

6. 成绩查询

1. 思路：计算及格人数以及优秀人数，再计算相关结果即可
2. 代码实现;

def search_grade():n = int(input())total = njige = 0great = 0for i in range(n):grade = int(input())if grade >= 85:jige += 1great += 1elif grade >= 60:jige += 1else:passprint(str(int(round(jige / n  * 100, 0))) + '%')print(str(int(round(great / n  * 100, 0))) + '%')return 0
search_grade()

4. 结果

7. 回文日期

1. 思路:考察对字符串的操作。在这里，我们直接对年份进行操作，之后翻转得到回文的月和日，判断月和日是否符合日期规则即可。
2. 代码实现:

from calendar import isleapgap_year = [0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
normal_year = [0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]def judge(date, is_gap):month = int(date[:2])day = int(date[2:])if month > 12:return Falseif is_gap:return day < gap_year[month]else:return day < normal_year[month]def judge_next_year(year: str):res_year = year[::-1]new_year = year + res_yearrevers_str = new_year[::-1]if new_year == revers_str:is_gap = isleap(int(year))flag = judge(res_year, is_gap)if flag:return [True, revers_str]else:return [False, revers_str]else:return [False, revers_str]def get_next(data: str):year = data[:4]res = list()while True:year = str(int(year) + 1)ans = judge_next_year(year)flag, tmp = ans[0], ans[1]if flag:res.append(tmp)breaknum = int(data[0:2])while True:num += 1year = str(num) + str(num)ans = judge_next_year(year)flag, tmp = ans[0], ans[1]if flag:res.append(tmp)breakreturn resif __name__ == '__main__':data = input()res = get_next(data)for i in res:print(i)

4. 结果：
   

8.子串分值和

1. 思路：动态规划，dp[i] = dp[i - 1] + change[i] （以s[i]结尾的字串的贡献为以前一个字符结尾的贡献 + s[i]的贡献）
2. 代码实现：

def son_str_sum(s):n = len(s)index_ls = [0] * 26index_ls[ord(s[0]) - 97] += 1dp = [0] * ndp[0] = 1for i in range(1, n):dp[i] = dp[i - 1] +  i + 1 - index_ls[ord(s[i]) - 97]index_ls[ord(s[i]) - 97] = i + 1return sum(dp)
if __name__ == '__main__':s = input()ans = son_str_sum(s)print(ans)

3. 答案：
   

9. 平面切分

1. 思路： 通过观察规律得到新加入的直线所分割成的平面数为1 + 加入直线之后平面中现存直线的交点
2. 代码实现：

def get_node(line1, line2):A1 = line1[0]B1 = line1[1]A2 = line2[0]B2 = line2[1]# 如果平行if A1 - A2 == 0:return Nonex = (B2 - B1) / (A1 - A2)y = A1 * x + B1# 避免浮点数计算中出现精度问题x = round(x, 8)y = round(y, 8)return (x, y)if __name__ == "__main__":n = int(input())lines = []for i in range(n):a, b = map(int, input().split())lines.append((a, b))lines = list(set(lines))n = len(lines)mid = [1] * (n + 1)node = set()for i in range(1, n):node.clear()for j in range(i):mux_node = get_node(lines[i], lines[j])if mux_node == None:passelse:node.add(mux_node)mid[i] +=  len(node)print(sum(mid))

3. 结果：
   

10.字串排序

1. 思路：暂时没想到，等之后想到了再更新