2021.07.17 ABC210 在线赛 一.比赛情况 组别 姓名 过题数量 1组 李涵 4 题 1组 王俊篪 5 题 1组 陈加哥 4 题 1组 禹芮霖 3 题 2组 刘洋 5 题 2组 董杰 4 题 2组 夏培轩 4 题 2组 余泽楷 3 题 3组 聂于涵 4 题 3组 肖放 3 题 3组 陈
考虑对于一个操作而言它做的贡献,这个题可以先不思考我们每个操作可以具体的连那些边,因为点的范围都到1e9了,显然是抽象的。u和v可以连边当且仅当u=(v+ai)u=(v+ai)%NN,换句话说可以写成u=v+ai+k∗Nu=v+ai+k∗N式子还可以转换成u=v+k∗gcd(ai,N)u=v+k∗gcd(ai,N)即u和v在模gcd(ai...
考虑对于一个操作而言它做的贡献,这个题可以先不思考我们每个操作可以具体的连那些边,因为点的范围都到1e9了,显然是抽象的。u和v可以连边当且仅当\(u=(v+a_i)\)%\(N\),换句话说可以写成\(u=v+a_i+k*N\)式子还可以转换成\(u=v+k*gcd(a_i,N)\)即\(u和v在模gcd(a_i,N)的意义上是同余...
A - Infinite Coins题目链接:https://abc088.contest.atcoder.jp/tasks/abc088_aTime limit : 2sec / Memory limit : 256MBScore: 1-yen coins and infinitely many N is an integer between A is an integer betwe 算法 代码 AtCoder Beginner Contest 087 (ABC) A - Buying Sweets题目链接:https:/...
abc210 D - National Railway DP arc002 C - コマンド入力 DP arc042 C - おやつ DP agc044 A - Pay to Win DP past201912-open I - 部品調達 DP past202004-open H - 1-9 Grid DP past202005-open H - ハードル走 DP abc037 C - 総和 累積和 abc089 D - Practical Skill Test...
AtCoder 的比赛分为ABC(AtCoder Beginner Contest),ARC(AtCoder Regular Contest) 以及 AGC (AtCoder ...
Contest URL: https://atcoder.jp/contests/abc332 Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20231210T2100&p1=248 Duration: 100 minutes Writer: math957963, leaf1415, evima, m_99 Tester: physics0523, cn449 Rated range: ~ 1999 The point values: 100-200-300-425...
283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 400 ABC_別解 AGC AHC ARC EDPC OTHERS memo 競プロ典型90題 精選100問 comfused.md readme.mdBreadcrumbs atcoder_codes /ABC /200 / 270/ Directory actions More options Failed to load latest commit information. ...
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 well on Codeforces because of system...
#include<bits/stdc++.h> #define int long long using namespace std; const int N=1e6+10; const int mod=998244353; int n,m; int a[210][210]; map<int,int>dp[210][210]; void solve(){ cin>>n>>m; for(int i=0;i<n;i++){ string s;cin>>s; for(int j=0;j<m;j++){ if...