} B - ASCII Art (abc294 b) 题目大意 给定一个二维数组,将其中的0变成.,其余的1∼261∼26则变成A∼ZA∼Z,输出变换后的数组。 解题思路 模拟即可。 神奇的代码 #include<bits/stdc++.h> usingnamespacestd; usingLL =longlong; intmain(void){ ios::sync_with_stdio(false); cin.tie(0); c...
I could not find the usual discussion post about the most recent atcoder contest, so I'm posting here. How to solve problem G?atcoder, beginner, contest, 294 +13 Ghassane 23 months ago 7 Comments (6) Show archived | Write comment? chromate00 23 months ago, # | +20 G: ...
1.AtCoder Beginner Contest 293(C,D ,E,F)2023-03-172.Educational Codeforces Round 115 (Rated for Div. 2)(D,E)2023-03-183.AtCoder Beginner Contest 294(E,F,G)2023-04-034.AtCoder Beginner Contest 2462023-03-295.中国石油大学(北京)第三届“骏码杯”程序设计竞赛(同步赛)(D,E,F)2023-03...
We will hold AtCoder Beginner Contest 284. Contest URL:https://atcoder.jp/contests/abc284 Start Time:http://www.timeanddate.com/worldclock/fixedtime.html?iso=20230107T2100&p1=248 Duration: 100 minutes Number of Tasks: 8 Writer:Nyaan,yuto1115 ...
AtCoder Beginner Contest 269 题解 A - E https://zhuanlan.zhihu.com/p/565544755 AtCoder Beginner Contest 269 (A-F)题解 https://www.cnblogs.com/ycllz/p/16704061.html AtCoder 刷题+题解 https://www.zhihu.com/column/c_1748429792083636225?page=2 ...
AtCoder Beginner Contest 359 A~F https://www.bilibili.com/video/BV1G4gveBENB/ AtCoder Beginner Contest 359 | 完成度 [6 / 7] https://www.bilibili.com/video/BV1kS411A7M3/ AtCoder Beginner Contest 359 A 至 G 題讲解 by dreamoon https://www.bilibili.com/video/BV1x1421r7gH/ AtCoder...
AtCoder Beginner Contest 042题解(ABCD) 传送门 A - Iroha and Haiku (ABC Edition) 签到题,直接判断是否是两个 5 5 5,一个 7 7 7。 #include<bits/stdc++.h> using namespace std; typedef long long ll; const int N=1e5+5,M=1e6+5,inf=0x3f3f3f3f,mod=1e9+7; ...
AtCoder Beginner Contest 044(ABCD)题解 传送门 A - Tak and Hotels (ABC Edit) 思路:显然讨论一下 k k k的范围即可。 时间复杂度: O ( 1 ) O(1) O(1) #include<bits/stdc++.h> using namespace std; typedef long long ll; ...
AtCoder Beginner Contest 296 B - Chessboard# 难度: ⭐# 题目大意# 给定一个8*8的字符矩阵, 其中只有一个' * ', 输出它的坐标; 其坐标的列用字母表示, 行用数字表示, 具体看样例解释; 解题思路# 签到题不多嗦了; 神秘代码# #include<bits/stdc++.h> #define int long long #define IOS ios::...
C - Merge Sequences 拼在一起后直接排序,在排序后数组上二分出位置即可。 #include <bits/stdc++.h> using namespace std; using ll = long long; void solve() { } int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int n, m; ...