2025-11-30 OJ题解
两道练手题目的题解和代码示例
宾果游戏
注意：字符串输入的处理、输出的格式要求
Bingo的条件包括：五行、五列、两条对角线任意一条全为'Y'即为Bingo
# 输入一个长度为5的字符串列表
network = [input() for _ in range(5)]
bingo = False
# 检查行
for row in network:
if row == "YYYYY":
bingo = True
# 检查列
for col in range(5):
if all(network[row][col] == "Y" for row in range(5)):
bingo = True
# 检查主对角线
if all(network[i][i] == "Y" for i in range(5)):
bingo = True
# 检查副对角线
if all(network[i][4 - i] == "Y" for i in range(5)):
bingo = True
# 三元表达式
print("Bingo" if bingo else "No")network = [input() for _ in range(5)]
bingo = False
# 检查行
for row in network:
if row == "YYYYY":
bingo = True
# 检查列
for col in range(5):
if network[0][col] == "Y" and network[1][col] == "Y" and network[2][col] == "Y" and network[3][col] == "Y" and network[4][col] == "Y":
bingo = True
# 检查主对角线
if network[0][0] == "Y" and network[1][1] == "Y" and network[2][2] == "Y" and network[3][3] == "Y" and network[4][4] == "Y":
bingo = True
# 检查副对角线
if network[0][4] == "Y" and network[1][3] == "Y" and network[2][2] == "Y" and network[3][1] == "Y" and network[4][0] == "Y":
bingo = True
if bingo:
print("Bingo")
else:
print("No")密码学
观察维吉尼亚密码的图片表格，可以发现每一行都是字母表的循环右移，我们要做的加密过程就是从明文字母所在的行开始，向右移动对应密钥字母个数的位置，得到密文字母。注意在移动的过程中，需要先将字母还原到0-25的数字区间进行计算，偏移后超出26对26取模，最后再转换回字母。
P = input()
K = input()
K = K * 114514
C = []
for p,k in zip(P, K):
print(chr((ord(p) - ord('A') + ord(k) - ord('A')) % 26 + ord('A')), end='')P = input()
K = input()
# 我不知道可以用K = K * (len(P) // len(K) + 1) 这种方式来扩展密钥字符串
# 我想到可以用字符串索引index来对密钥长度取模
C = []
for i in range(len(P)):
k = K[i % len(K)]
p = P[i]
# 密钥相对于A的偏移量
offset = ord(k) - ord('A')
# 加密
c = chr((ord(p) - ord('A') + offset) % 26 + ord('A'))
# 为什么明文先减去'A'?
# 字母表是26一循环 所以要把字母映射到0-25之间
# 我不知道可以print(chr(...), end='') 这种方式直接输出密文
C.append(c)
# 使用''.join()来拼接字符串列表
print(''.join(C))