Arpa's loud Owf and Mehrdad's evil plan time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output As you have noticed, there are lovely girls in Arpa's land. People in Arpa's land are numbered from 1…

D. Arpa's weak amphitheater and Mehrdad's valuable Hoses Problem Description: Mehrdad wants to invite some Hoses to the palace for a dancing party. Each Hos has some weight wi and some beauty bi. Also each Hos may have some friends. Hoses are divided…

ARPA是英文Advanced Research Projects Agency的缩写,代表美国国防部高级研究计划署.是美国国防部高级研究计划管理局因军事目的而建立的,开始时只连接了4台主机,这便是只有四个网点的网络之父…

Arpa's weak amphitheater and Mehrdad's valuable Hoses time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Just to remind, girls in Arpa's land are really nice. Mehrdad wants to invite some Hose…

Arpa's loud Owf and Mehrdad's evil plan time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output As you have noticed, there are lovely girls in Arpa’s land. People in Arpa's land are numbered from 1…

C. Arpa's loud Owf and Mehrdad's evil plan time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output As you have noticed, there are lovely girls in Arpa’s land. People in Arpa's land are numbered fro…

B. Arpa’s obvious problem and Mehrdad’s terrible solution time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output There are some beautiful girls in Arpa’s land as mentioned before. Once Arpa came u…

题意:给定一个正整数序列,两人轮流对这个数列进行如下修改:选取一个素数p和一个整数k将序列中能整除p^k的数除以p^k,问谁有必胜策略. 借此复习一下sg函数吧,sg(x) = mex ( sg(y) |y是x的后继结点 ).我们不难发现不同的质因子是互不影响的,因此我们可以把不同的质因子归为不同的game.因为每次操作对整个序列有效,所以序列中p^k的个数也是不影响答案的.因此我们可以用一个二进制位表示当前序列是否存在p^k,如果存在,则其第(k-1)位为1.由是把所有game的sg异或起来即…

D. Arpa's letter-marked tree and Mehrdad's Dokhtar-kosh paths CF741D 题意: 一棵有根树,边上有字母a~v,求每个子树中最长的边,满足这个边上的所有字母重拍后可以构成回文 发明者自己出的题...orz 由于本来知道就是dsu on tree,所以还是想出来了 首先点分治是没法做了,这是有根树 写成二进制,两条链合起来构成回文\(\rightarrow\)异或和为0或者只有一位是1 一开始困惑于只处理到当前根的异或和的话,随着当前…

首先容易想到,每种素数是独立的,相互sg就行了 对于一种素数来说,按照的朴素的mex没法做... 所以题解的简化就是数位化 多个数同时含有的满参数因子由于在博弈中一同变化的,让他们等于相当于,那么这样就是一个数了 之后就是模拟,牛逼的思路 #include<iostream> #include<map> #include<iostream> #include<cstring> #include<cstdio> #include<set>…