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个人简介
茶亦醉人何必酒,书能香我不须花
本用户已离线......
该用户很懒,这里啥也没写用户不懒,所以留下了什么......
$\small\texttt{\color{#FA4129}本}\huge\texttt{\color{#FE9019}人}_{\small\texttt{\color{#FFE304}的}^{\large\texttt{\color{#FFEC01}萌\color{#FFF900}新}\small\texttt{\color{#FCFB03}Q\color{#F8FB07}W\color{#F1FB0B}Q}}}^{\large\texttt{\color{#FFB511}是}{\small\texttt{\color{#FFDC07}刚\color{#FFEF00}学}\large\texttt{\color{#FFF600}O\color{#FFFA00}I}}}\huge\texttt{\color{#E6F911}但\color{#92E82F}是}^{\large\texttt{\color{#39D54B}即}{\small\texttt{\color{#03C767}使}}}_{\normalsize\text{\color{#07C964}是\color{#00C789}这\color{#00C7A5}样}}\texttt{\color{#00CBC6}我\color{#00D0EB}也}^{\small\texttt{\color{#00D0F2}要}\normalsize\texttt{\color{#00D0F6}用}\texttt{\color{#03BEF4}蒟}_{\texttt{\color{#04AAEF}蒻}\large\texttt{\color{#078DE4}的}}}_{\scriptsize\texttt{\color{#01CDF6}声\color{#03C2F5}音\color{#04B4F2}大\color{#04A7EE}声\color{#0791E6}喊\color{#0A7BDD}出}}\mathcal{\color{#125BCD}I\color{#3D2AB5}AK\color{#A011AD}IOI!}$
我的账户
${\color{Orange} \mathbf{\mathcal{ Code Forces} } } $
。。。
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我的留言板(请不要在留言板里瞎写东西)
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本人在里最高
本人在里最高
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当你想离开时,你犹豫了
往下看
....... ......
用户不是很懒了,所以用户也不是蒟蒻
该用户很喜欢刷题,可是往往题却对我很不友好
题库
<!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8"> <title>Title</title> </head> <body> <h1>Hello World</h1> <h2> <a href="http://www.baidu.com"> My name is Wonder. </a> </h2> <input type="date"> <button>确定</button> <h3>全是瞎写的,不要介意<h3> </body> </html>
# 简单的快速排序,从大到小 def quick_sort(number_list): if len(number_list) <= 1: return number_list pivot = number_list[0] left = [] for i in range(1, len(number_list)): if number_list[i] > pivot: left.append(number_list[i]) right = [] for i in range(1, len(number_list)): if number_list[i] <= pivot: right.append(number_list[i]) return quick_sort(left) + [pivot] + quick_sort(right) number_list = input('直接填数字,不用空格') print(quick_sort(number_list))
// 简单的蛇形方阵 # include <bits/stdc++.h> using namespace std; int a[15][15]; int main() { int n; cin >> n; int k = 1; int x = 0; int y = 0; a[x][y] = k; k++; while (k <= n * n) { while (y + 1 < n && a[x][y + 1] == 0) { y += 1; a[x][y] = k; k++; } while (x + 1 < n && a[x + 1][y] == 0) { x += 1; a[x][y] = k; k++; } while (y - 1 >= 0 && a[x][y - 1] == 0) { y -= 1; a[x][y] = k; k++; } while (x - 1 >= 0 && a[x - 1][y] == 0) { x -= 1; a[x][y] = k; k++; } } for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { printf("%3d", a[i][j]); } printf("\n"); } return 0; }
目前就更新到这儿
用户太懒了,但用户也不是蒟蒻
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