ABC(13) 动态规划(8) 线段树(5) 贪心(5) 编译原理(5) LIS(2) LCA(2) dp(2) dfs(2) 更多 随笔分类 (104) ABC(14) C++(6) Codeforces(5) Java学习(1) LeetCode周赛(31) OS复习(9) 编译原理(5) 计算机系统结构(5) 算法(13) 算法设计与分析复习(14) 学习(...
AtCoder ABC236E. Average and Median(青色) 原题链接中等 作者: 白流雪 , 2023-05-17 22:45:47 , 所有人可见 , 阅读 161 2 1 算法(二分答案、动态规划)对于第一个问题:可以对答案进行二分 能否从 AiAi 中选出几个数使得平均值 ⩾x⩾x ⇔⇔ 能否从 Ai−xAi−x 中选出几个数满足...
时间复杂度 \(O(n)\)。 ABC236Ex Distinct Multiples (2900) 显然唯一难搞的就是 “互不相同” 条件,直接考虑容斥。 也许会考虑钦定一些点填的数必须相同,但这样需要记录若干组点集,难以实现。 不妨考虑钦定边集(两个点填的数相同即连边)。设 \(f_S\) 表示恰好连 \(S\) 边的方案数,\(g_S\) 表示...
Generated testcase example for N=8 A 5.6% failure chance per testcase isn't very encouraging, though it's unlikely that all 28 testscases from the AtCoder's set are stressing the worst possible scenario. So I got my AC verdict on the 3rd attempt during the contest. This wouldn't work...
They also did the similar thing in ABC354. This time (ABC356), there are much more suspicious participants. The average first-AC time (in seconds) of ABC347~ABC353 is in the following table: ContestABCDEFG ABC347 21 45 93 244 177 300 824 ABC348 15 37 72 302 236 495 430 ABC349 ...
https://atcoder.jp/contests/abc335/tasks/abc335_d 代码分享 代码语言:javascript 代码运行次数:0 运行 AI代码解释 #include <bits/stdc++.h> using namespace std; int g[50][50], a; void color(int start, int m) { for (int i = start; i < m; ++i) { g[start][i] = a; ++a; ...
ABC236-D DanceE - NAND repeatedly問題ページ考察動的計画法DPで解きます。 例えば N=3 だったとして、すべての f(l,r) の値が分かっているとします。 このときの数列の末尾に 0 か1 を追加したときに、答えがどう変わるかを考えます。 答えには f(1,4)+f(2,4)+f(3,4)+f(4,...
abc250 D - 250-like Number 素数 abc215 D - Coprime 2 素因数分解 abc233 C - Product DFS abc236 D - Dance DFS abc213 D - Takahashi Tour DFS abc284 E - Count Simple Paths DFS abc240 E - Ranges on Tree DFS abc239 E - Subtree K-th Max DFS abc014 D - 閉路 Euler Tour, Segment...
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string a[4] = {"ARC","AGC","ABC","AHC"}; int b[4]= {0}; for(int i= 0 ; i < 3 ;i++) { string t; cin >> t; for(int i = 0 ; i < 4 ;i++)if(a[i] == t)b[i] = 1; } for(int i = 0 ; i < 4; i++)if(!b[i])cout<<a[i]<<'\n'; C...