· AtCoder Beginner Contest 236(D,E,F) · AtCoder Beginner Contest 283(A~F) 阅读排行: · 开发的设计和重构,为开发效率服务 · 从零开始开发一个 MCP Server! · .NET 原生驾驭 AI 新基建实战系列(一):向量数据库的应用与畅想 · ThreeJs-16智慧城市项目(重磅以及未来发展ai) · Ai满嘴...
AtCoder Beginner Contest 236 题解 A,B,CA,B,C题过水,不作赘述,我的AtcoderAtcoder号是ComplexComplex,需要代码可以去找找。 DD题DanceDance 把2n2n个人按每组两人分成nn组,第ii人和第jj人同一组会产生Ai,jAi,j的贡献,求nn组贡献值的异或和最大值。
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...
AtCoder Beginner Contest 269(E-EX) https://www.bilibili.com/opus/707445585544740886 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 Beginner Contest 174 题解 AtCoder Beginner Contest 174 题解C Repsept 知识点:取模对加法和乘法封闭,所以暴力到2×1062\times 10^62×106,一边取模一边特判即可。D - Alter Altar 最终肯定是RRR全在左边,简单证明:假设最终RRR左边的某个位置存在WWW,则该位置右边只能是WWW,然后就是子问题了...
AtCoder Beginner Contest 291 | ABC 吊打萌新珍贵录像 | A~G 题解 121 -- 2:34:14 App yukicoder contest 435 | 完成度 [5 / 7] 274 -- 1:56:52 App AtCoder Beginner Contest 363 | 完成度 [6 / 7] 129 -- 2:34:36 App Codeforces Round 955 (Div. 2) | 完成度 [4 / 6] 354...
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...
We will hold AtCoder Beginner Contest 167. Contest URL: https://atcoder.jp/contests/abc167 Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20200510T2100&p1=248 Duration: 100 minutes Number of Tasks: 6 Writer: gazelle, kort0n, kyopro_friends, potetisensei, she...
AtCoder Beginner Contest 399(C-F详细题解) C 思路:作为一个简单无向图,有 n 个点只能有 n-1 条边,那我们只需要统计不同连通图上的点计算出所需的边即可,最后就是总边数减去所需的边即可。 代码:vector<vector<int>>… 秋日薄雾 AtCoder Beginner Contest 400 A-F 简易题解,如果题解...
头一次写题解。 A - A Healthy Breakfast r在m前面输出Yes,反之输出No。 统计两个位置比一下就行。 voidsolve(){strings;cin>>s;inta=0,b=0;for(inti=0;i<3;i++)if(s[i]=='R')a=i;elseif(s[i]=='M')b=i;if(a<b)yes;elseno;} ...