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题解 | 点击消除

s = input() stack = [] for i in s: if not stack : stack.append(i) else : if stack[-1] == i : stack.pop(-1) ...

2025-12-09

0 5

题解 | 小红的异或构造

n = int(input()) C = n+1 ans = [C ^ i for i in range(1, n + 1)] print(' '.join(map(str, ans))) 本题需要用到异或的数学性质，不难发现本题的构造可以改写为ai异或i = aj 异或j，这就表明数列的所有异或结...

2025-12-07

0 12

题解 | 构造序列

n,m = map(int,input().split()) if m==n: print(m*2) elif m==0 or n==0: print(1) else : print(min(n,m)*2+1)

2025-12-06

0 9

题解 | 如果直接枚举所有的肯定会超时，所以要先预处理记录#的位置

n = int(input()) list0 = [list(input().strip()) for _ in range(n)] # 收集所有#的位置 points = [] for i in range(n): for j in range(n): if list0[i...

2025-12-06

0 8

题解 | 简单的常规模拟

n = int(input()) s = input().split() new = "" for i in range(len(s)): if s[i] == "0" : continue else : p =...

2025-12-06

0 8

题解 | 铺地毯

import sys n = int(input()) list0 = [list(map(int, input().split())) for i in range(n)] x,y = map(int, input().split()) t = -1 for j in range(n): ...

2025-12-02

0 9

题解 | 排序危机之利用py的异常机制的一个很邪门的解法

#很不推荐大家在正赛用try，有点不规范，但可以开阔下思路（ a = int(input()) n = input() low = "" up = "" num = "" for i in n: try : p = ...

2025-12-02

1 11

题解 | 添加逗号

n = input() m = n[::-1] count = 0 s = "" for i in m: count += 1 if count % 3 == 0 and count != len(m): s += i + ","...

2025-12-02

0 10