XXII Open Cup. Grand Prix of Korea will be held in 2021/10/24 Sunday, 17:00 KST (UTC+9). For external accounts, the contest is ready now. List of relevant previous contests: XVIII Open Cup. GP of Korea KAIST RUN Spring Contest 2018 XIX Open Cup. GP of Korea XIX Open Cup. GP...
考虑第一个点的出栈点设其为xx,在此时栈为空,所以相当于拆成了两个子问题,其中要满足S2..xS与T1..x−1T0/1 个数相等。那么同理,第二个点就是在第一个出栈点的左边再找一个满足此类似条件的出栈点。容易发现,我们贪心让出栈点靠右是正确的,所以双指针扫一下就行了。
XXII Open Cup. Grand Prix of Korea 目录 H.Or Machine J.Periodic Ruler A.Automatic Sprayer 2 H.Or Machine 4s Problem 给定nn 个不超过 2828 的数字 x1 xnx1 xn。 给定ll 个操作 (a,b)(a,b),表示 xa=xa|xbxa=xa|xb,其中 || 是按位或。 给定tt 表示进行 tt 次操作,...
Is the opencup website not available outside of contests? I haven't been able to load it in quite some time. → Reply wilcot 2 years ago, # | ← Rev. 4 +23 Editorial for D: Note that the rotation operation in a binary tree does not change the order of numbers during forward...
Gym103371(XXII Open Cup. Grand Prix of Korea) A. Automatic Sprayer 2 只能膜拜孔姥爷,换我肯定找不全方程。 设每一行的和为 \(R_i\),每一列的和为 \(C_i\),那么有: \[E_{x,y} = \sum_{i=1}^{n}|x-i|R_i+ \sum_{j=1}^{n} |y-j|C_j \]...
XXII Open Cup. Grand Prix of Korea H.Or Machine 4s Problem 给定\(n\)个不超过\(2^8\)的数字\(x_1~x_n\)。 给定\(l\)个操作\((a,b)\),表示\(x_a = x_a | x_b\),其中\(|\)是按位或。 给定\(t\)表示进行\(t\)次操作,按照操作\(0\),操作\(1\),……,操作\(l-1\),...
输入n为负数,从n+1开始找可行解即可 输入n为正数,若有可行解则输出,等价输入负数处理 E. Median 要保证中位数为珍贵数,只需要保证连续的不珍贵数长度len * 2小于不珍贵数总长度sum,或者 小于不珍贵数总长度和以及珍贵数之和 关键代码 for(inti=1;i<=n&&!flag;i++){if(a[i]){cnt+=a[i];continue...
XXII Open Cup. Grand Prix of Seoul 部分题解 比赛链接:https://codeforc.es/gym/103855。 A. Factory Balls 题意 给出nn个编号为[1,n][1,n]的点,初始每个点的颜色都为11。有mm种装备,每种装备会遮住一些位置(不一定是连续区间)。每一个时刻可以进行如下操作之一:...
XXII Open Cup. Grand Prix of Korea 链接ABCDEFGHIJKLM +1+1 +3+3 +1+1 ++ ++ ++ +1+1 ++ ++ J. Periodic Ruler签到题。H. Or Machine对于整周期我们按位处理,记录一个数字至少多少轮后该位变成 11。这本质上是一个最短路,而且有效转移边边权只有 00 和11。
You run it for C<=1e4 and see that you only need small number of B (200ish). IIRC For C <= 2e5 ans <= 301 → Reply tfg 3 years ago, # ^ | +18 I did that + realize that the pattern depends on the parity of N, so just make a program to print all ranges of ...