Codeforces Round #588 (Div. 2)-E. Kamil and Making a Stream-求树上同一直径上两两节点之间gcd的和 [Problem Description] 给你一棵树,树上每个节点都有一个权值.定义\(1\sim v\)的最短路径所经过的所有节点\(u\)称为\(v\)节点的祖先.定义函数\(f(u,v)=gcd(u,t1,t2,\dots,v)\),其中\(u,t1,t2,\dots\)都是\(v\)的祖先.求\(\sum f(u,v)\). […

链接: https://codeforces.com/contest/1230/problem/E 题意: Kamil likes streaming the competitive programming videos. His MeTube channel has recently reached 100 million subscribers. In order to celebrate this, he posted a video with an interesting problem…

链接: https://codeforces.com/contest/1230/problem/D 题意: Marcin is a coach in his university. There are n students who want to attend a training camp. Marcin is a smart coach, so he wants to send only the students that can work calmly with each other. L…

链接: https://codeforces.com/contest/1230/problem/C 题意: Anadi has a set of dominoes. Every domino has two parts, and each part contains some dots. For every a and b such that 1≤a≤b≤6, there is exactly one domino with a dots on one half and b dots on th…

链接: https://codeforces.com/contest/1230/problem/B 题意: Ania has a large integer S. Its decimal representation has length n and doesn't contain any leading zeroes. Ania is allowed to change at most k digits of S. She wants to do it in such a way that S…

链接: https://codeforces.com/contest/1230/problem/A 题意: Dawid has four bags of candies. The i-th of them contains ai candies. Also, Dawid has two friends. He wants to give each bag to one of his two friends. Is it possible to distribute the bags in suc…

传送门 A. Dawid and Bags of Candies 乱搞. Code #include <bits/stdc++.h> #define MP make_pair #define fi first #define se second #define sz(x) (int)(x).size() //#define Local using namespace std; typedef long long ll; typedef pair<int, int> pii; con…

Contest Page 因为一些特殊的原因所以更得不是很及时-- A sol 不难发现当某个人diss其他所有人的时候就一定要被删掉. 维护一下每个人会diss多少个人,当diss的人数等于剩余人数$-1$的时候放队列里,每一次取队头更新其他人diss的人数. code B sol 一个结论:对于序列$a_1,a_2,...,a_n$,其前缀$gcd$数量不超过$log_2a_i$种.证明考虑从前往后计算前缀$gcd$,那么从第$i-1$个$gcd$到第$i$个$gcd$,数值要么不变要么至少…

1. 1229A Marcin and Training Camp 大意: 给定$n$个对$(a_i,b_i)$, 要求选出一个集合, 使得不存在一个元素好于集合中其他所有元素. 若$a_i$的二进制中某一位为$1$, $a_j$对应位为$0$, 那么$(a_i,b_i)$好于$(a_j,b_j)$. 显然出现次数不少于$2$的数都可以选, 然后再把包含关系的数全选了即可. #include <iostream> #include <sstream> #include <al…

思路:先找出现次数>=2数.然后在取跑所有数,需要考虑一般情况(当一个人比另一个人的ai小且他们的与运算等于小的那个人的ai那么可以知道大的那个人必定强于ai小的那个人). 则可以用位运算实现判断强弱. #include<bits/stdc++.h> using namespace std; #define int long long #define N 100500 struct str{ int a,b; }st[N]; map<int,int> mp; vector&l…