C. Alternating Subsequence time limit per test 1 second memory limit per test 256 megabytes Recall that the sequencebbis a a subsequence of the sequenceaaifbbcan be derived fromaaby removing zero or more elements without changing the order of the remaining elements. For example, ifa=[1,2,1...
C. Even Path time limit per test 1 second memory limit per test 256 megabytes Pathfinding is a task of finding a route between two points. It often appears in many problems. For example, in a GPS navigation software where a driver can query for a suggested route, or in a robot motion...
Problem - 433C - Codeforces解题报告 对于这题本人刚开始的时候的想法是:先把最大两数差的位置找到然后merge计算一个值再与一连串相同的数做merge后计算一个值比较取最大值输出;可提交后发现不对,于是本人就搜了一下正解发现原来这题的正确解题思路是:采用数学中的中位数原理,分别把某数两边相邻且不同的数存...
Problem Link A. Almost Prime time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A. Almost P
题目链接:Problem - E - Codeforces 题意: t组数据,给定n个点、点u有颜色au的有根树,点1为该树的根,定义f(u,v)为u→v的路径上的颜色数目,u不是v祖先时无定义。对于所有点对(u,v),其中u不一定相异于v,设l为u,v的最近公共祖先,求maxf(l,u)⋅f(l...
Codeforces Beta Round #2 C. Commentator problem 模拟退火果然是一个非常高端的东西,思路神马的全然搞不懂啊~ 题目大意: 给出三个圆,求一点到这三个圆的两切线的夹角相等。 解题思路: 对于这个题来说还是有多种思路的 。只是都搞不明确~~ /害羞脸...
Problem C. Painting Cottages Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100342/attachments Description The new cottage settlement is organized near the capital of Flatland. The construction company that is building the settlement has decided to paint some cottages ...
Repository files navigation README codeforces solutions Codeforces problem solutions My handle is : http://codeforces.com/profile/misra.ji Please do notify me @ joyneel.misra@students.iiit.ac.in for any queries or issues :)About codeforces problem solutions Resources Readme Activity Stars 2...
Codeforces gym/102956/problem/F 前oier,entp摩羯 2020-2021 Winter Petrozavodsk Camp, Belarusian SU Contest (XXI Open Cup, Grand Prix of Belarus) F. Border Similarity Undertaking 本来自己总结完不太想写的,但是想这道题过程一开始只想到了bitset的做法或者分治但是不太会合并那个玩意,还是记录一下。感谢...
你的算法需要优化。其实不需要去找出每对匹配的i j，只要找出数量就可以了。我给段简单的代码，你看看应该就明白了。include <stdio.h> double table[256];char str[100001];void main(){ char *p = str;int i;double n=0;gets(str);while(*p)table[*(p++)]+=1;for (int i = 1; i...