time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output Kevin Sun has just finished competing in Codeforces Round #334! The round was 120 minutes long and featured five problems with maximum point valu…
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=102271#problem/B Description Kevin Sun has just finished competing in Codeforces Round #334! The round was 120 minutes long and featured five problems with maximum point values of 500, 1000, 1500,…
题目链接 可以发现 十进制4 对应 二进制100 十进制16 对应 二进制10000 十进制64 对应 二进制1000000 可以发现每多两个零,4的次幂就增加1. 用string读入题目给定的二进制数字,求出其长len,当len为奇数时,第一位为1,后面的位数如果都为0,则输出len,如果有一个不为0,则输出len+1: 当len为偶数时,则输出len.(之所以这样输出是因为题目给定4的次幂是从0开始的) #include<iostream> #include<string> #…
time limit per test3 seconds memory limit per test256 megabytes inputstandard input outputstandard output Anton is growing a tree in his garden. In case you forgot, the tree is a connected acyclic undirected graph. There are n vertices in the tree, e…
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output Generous sponsors of the olympiad in which Chloe and Vladik took part allowed all the participants to choose a prize for them on their own. Chri…
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output Ari the monster is not an ordinary monster. She is the hidden identity of Super M, the Byteforces' superhero. Byteforces is a country that consi…
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output Spongebob is already tired trying to reason his weird actions and calculations, so he simply asked you to find all pairs of n and m, such that t…
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output Vika has an infinite sheet of squared paper. Initially all squares are white. She introduced a two-dimensional coordinate system on this sheet a…
[链接] 我是链接,点我呀:) [题意] [题解] 先处理出来任意一棵树. 然后把不是树上的边处理出来 对于每一条非树边的点(最多21*2个点) 在原图上,做dijkstra 这样就能处理出来这些非树边上的点到其他任意点的最短路了. 然后对于询问x,y 先用LCA+预处理,求出树上的最短路. 接下来考虑有非树边的情况. 显然只要枚举它经过了非树边上的点z 那么用dis[z][x]+dis[z][y]尝试更新ans就好. 只要枚举非树边上的点. 这是突破口. [代码] #include <bits…
time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by…