Codeforces Round 940 (Div. 2) and CodeCraft-23Codeforces Round 940 (Div. 2) and CodeCraft-23 个人写题记录 2024-04-27 ACM&算法 #ACM #Codeforces
Educational Codeforces Round 161 (Rated for Div. 2)Educational Codeforces Round 161 (Rated for Div. 2) 个人写题记录 2024-03-22 ACM&算法 #ACM #Codeforces
Pinely Round 3 (Div. 1 + Div. 2)Pinely Round 3 (Div. 1 + Div. 2) 个人写题记录 2024-02-24 ACM&算法 #ACM #Codeforces
Educational Codeforces Round 160 (Rated for Div. 2)Educational Codeforces Round 160 (Rated for Div. 2) 个人写题记录 2024-02-17 ACM&算法 #ACM #Codeforces
Educational Codeforces Round 159 (Rated for Div. 2)Educational Codeforces Round 159 (Rated for Div. 2) 个人写题记录 2024-02-13 ACM&算法 #ACM #Codeforces
CodeTON Round 7 (Div. 1 + Div. 2, Rated, Prizes!)CodeTON Round 7 (Div. 1 + Div. 2, Rated, Prizes!) 个人写题记录 2024-01-28 ACM&算法 #ACM #Codeforces
Educational Codeforces Round 158 (Rated for Div. 2)Educational Codeforces Round 158 (Rated for Div. 2) 个人写题记录 2024-01-19 ACM&算法 #ACM #Codeforces
Educational Codeforces Round 157 (Rated for Div. 2)Educational Codeforces Round 157 (Rated for Div. 2) 个人写题记录 2024-01-01 ACM&算法 #ACM #Codeforces
反复横跳的 Clang-Tidy(cert-dcl21-cpp)Clang-Tidy 在 operator++(int) 方法的返回值是否要加 const 这件事反复横跳 2023-12-04 杂项 #C++
Codeforces Round 904 (Div. 2)D 题有点难,数论确实不会,本着只是为了练习回复脑子的角度考虑,就不写了 A. Simple Design大致题意有两值,$x, k$,找到最小的 $y$ 满足 $y \geq x, y \space mod \space k = 0$ 思路因为 $k$ 很小,所以暴力枚举就行 AC code1234567891011121314151617181920212223242526void solve 2023-11-25 ACM&算法 #ACM #Codeforces
Codeforces Round 903 (Div. 3)A. Don’t Try to Count大致题意给出两个字符串 $n, m$,允许 $n$ 每次往自己拼接在自己后面,使得 $n$ 中出现 $m$ 字符串,问最少需要几次操作 思路因为 $n, m$ 都很小,所以直接保留就行了 AC code1234567891011121314151617181920void solve() { int _; cin >> _ 2023-11-21 ACM&算法 #ACM #Codeforces