VP AtCoder Beginner Contest 390 A - 12435点击查看代码 void solve() { int cnt = 0; int a[6]{}; for (int i = 1; i <= 5; ++ i) { std::cin >> a[i]; } for (int i = 1; i <= 4; ++ i) { cnt += (a[i] != i && a[i + 1...
【Atcoder训练记录】AtCoder Beginner Contest 390 训练情况赛后反思A题犯蠢冒泡排序写错,B题等比数列不能使用除法算公比,会有误差问题,成比例要用乘法比较A题我们观察到 12345 是有序的,所以我们对于给出的数列进行冒泡排序,次数为 1 次答案就为 YES 否则为 NO...
AtCoder Beginner Contest 403 A-G 简易题解,如果题解中有什么问题可以找我反馈,谢谢! A.Odd Position Sum直接 for 循环输入,只对奇数求和即可。int main(){ int n; cin >> n; int res = 0; for(int i = 1; i… 枫落发表于Atcod... AtCoder Beginner Contest 401 A-G 简易题解,如果...
push_back({a, c}); } vector<vector<int>> dp(3, vector<int>(N, -INF)); for (int v = 0; v < 3; v++) { dp[v][0] = 0; for (auto [a, c] : arr[v]) { for (int i = x; i >= c; i--) { dp[v][i] = max(dp[v][i], dp[v][i - c] + a); } for ...
判断图中存在闭环的常用方法——以Atcoder Beginner Contest 285(D - Change Usernames)为例 737 0 30:29 App AtCoder Beginner Contest 392(A ~ G 题讲解) 559 0 21:08 App Atcoder Beginner Contest 338 289 0 17:35 App Atcoder Beginner Contest 348(A ~ F 讲解,无 D) 1326 0 22:18 App ...
·AtCoder Beginner Contest (ABC) 这是最频繁且最简单的入门赛,通常情况下每月至少举行2次。2019年4月27日(含)之前,每场比赛共4题,时长100分钟,满分1000分且Rating超过1199的选手不计Rating值。自2019年5月19日起改版升级为6道题目,时长不变,满分2100分且Rating值超过1999的选手不计Rating值。改版之后比赛质量...
Case 1 is pretty simple : if all these 3 parts give the same answer , then to reverse the max appearing element , you need to take the 2 minimum counts from these 3 (counts means the min number of swaps required in those subparts , i.e we are returning a pair<int,int> in dp ,...
We will hold AtCoder Beginner Contest 396. Contest URL: https://atcoder.jp/contests/abc396 Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20250308T2100&p1=248 Duration: 100 minutes Writer: toam, sounansya Tester: physics0523, yuto1115 Rated range: ~ 1999 The point...
G题在罚坐,剪掉了。, 视频播放量 212、弹幕量 0、点赞数 4、投硬币枚数 4、收藏人数 4、转发人数 0, 视频作者 CreatiQ, 作者简介 Arcaea/ICPC 区域赛铜/SCNU/大一/数学与应用数学(师范),相关视频:AtCoder Beginner Contest 399 | D E F 讲解,【坐牢2h13min】Teza Roun
AtCoder Beginner Contest 193 部分题解 E - Oversleeping 求是否存在\(t\)满足\(t=t_1(mod (2X+2Y)) and t=t_2(mod (P+Q))\) 注意到\(Q\)和\(Y\)非常小,直接枚举套个\(exCRT\)就行了(虽然赛场上没看出来,\(exCRT\)也忘了记得快速乘...