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…

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…

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…

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…

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…

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…

http://www.cnblogs.com/a2985812043/p/7224574.html 解法:这是网上看到的 因为要计算u->v的权值之和,我们可以把权值放在v中,由于题目中给定的u.v特性,我们可以从最后一个v开始倒回来每次除以2,然后把权值加起来就好了,注意输入的区间大小值 #include <iostream> #include <cstdio> #include <map> using namespace std; #define LL lon…

方法:求出最近公共祖先,使用map给他们计数,注意深度的求法. 代码如下: #include<iostream> #include<cstdio> #include<map> #include<cstring> using namespace std; #define LL long long map<LL,LL> sum; int Get_Deep(LL x) { ; i < ; i++) { ))>x) ; } ; } void…