C. Lorenzo Von Matterhorn time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Barney lives in NYC. NYC has infinite number of intersections numbered with positive integers starting from 1. Ther…

2018-03-16 http://codeforces.com/problemset/problem/697/C C. Lorenzo Von Matterhorn time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Barney lives in NYC. NYC has infinite number of intersect…

Lorenzo Von Matterhorn Barney lives in NYC. NYC has infinite number of intersections numbered with positive integers starting from 1. There exists a bidirectional road between intersections i and 2i and another road between i and 2i + 1 for every pos…

Lorenzo Von Matterhorn time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Barney lives in NYC. NYC has infinite number of intersections numbered with positive integers starting from 1. There e…

题目链接: A. Lorenzo Von Matterhorn time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Barney lives in NYC. NYC has infinite number of intersections numbered with positive integers starting from 1…

http://codeforces.com/contest/697/problem/C C. Lorenzo Von Matterhorn time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Barney lives in NYC. NYC has infinite number of intersections numbered…

原题链接:http://codeforces.com/contest/696/problem/A 原题描述: Lorenzo Von Matterhorn   Barney lives in NYC. NYC has infinite number of intersections numbered with positive integers starting from 1. There exists a bidirectional road between intersections i a…

A. Lorenzo Von Matterhorn time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Barney lives in NYC. NYC has infinite number of intersections numbered with positive integers starting from 1. Ther…

Least Common Ancestors 节点范围是1~1e18,至多1000次询问. 只要不断让深的节点退一层(>>1)就能到达LCA. 用点来存边权,用map储存节点和父亲连边的权值. #include<cstdio> #include<map> #define ll long long using namespace std; map<ll,ll>m; ll u,v,w; void add(){ while(u!=v){ if(u<v){ m…

题目链接:http://codeforces.com/problemset/problem/697/D 给你一个有规则的二叉树,大概有1e18个点. 有两种操作:1操作是将u到v上的路径加上w,2操作是求u到v上的路径和. 我们可以看出任意一个点到1节点的边个数不会超过64(差不多就是log2(1e18)),所以可以找最近相同祖节点的方式写. 用一条边的一个唯一的端点作为边的编号(比如1到2,那2就为这条边的编号),由于数很大,所以用map来存. 进行1操作的时候就是暴力加w至u到LCA(u,v…