Space Golf 题目连接: http://codeforces.com/gym/100803/attachments Description You surely have never heard of this new planet surface exploration scheme, as it is being carried out in a project with utmost secrecy. The scheme is expected to cut costs of c…

B. Island Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100114 Description On February 30th this year astronauts from the International Space Station flew over the Pacific Ocean and took a picture, on which was discovered a pr…

原题链接:http://codeforces.com/gym/100431/attachments/download/2421/20092010-winter-petrozavodsk-camp-andrew-stankevich-contest-37-asc-37-en.pdf 题意 神奇的电路可以通过比较器来实现冒泡. 题解 倒着往回考虑,搞搞就好. 代码 #include<iostream> #include<cstdio> #include<cstring> #…

C. CinderellaTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100685/problem/C Description Cinderella is given a task by her Stepmother before she is allowed to go to the Ball. There are N (1 ≤ N ≤ 1000) bottles with water in th…

Description You surely have never heard of this new planet surface exploration scheme, as it is being carried out in a project with utmost secrecy. The scheme is expected to cut costs of conventional rover-type mobile explorers considerably, using pr…

Problem G.Giant ScreenTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=86821#problem/G Description You are working in Advanced Computer Monitors (ACM), Inc. The company is building and selling giant c…

题意简述 给定一个由A C G T四个字母组成的密码锁(每拨动一次 A变C C变G G变T T变A) 密码锁有n位 规定每次操作可以选取连续的一段拨动1~3次 问最少几次操作可以将初始状态变到末状态 并且把每次操作输出 (此题有spj) --------------------------------------------------------------------------------------------------------- 为了方便叙述 所有没有明确说明的位置均是在$mod…

A. Wizards and Trolleybuses Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/167/problem/A Description In some country live wizards. They love to ride trolleybuses. A city in this country has a trolleybus depot with n trolle…

一句话题意:x0=1,xi+1=(Axi+xi%B)%C,如果x序列中存在最早的两个相同的元素,输出第二次出现的位置,若在2e7内无解则输出-1. 题解:都不到100天就AFO了才来学这floyd判圈算法. 介绍一下floyd判圈算法:该算法适用于在线性时间复杂度内判断有限自动机.迭代函数.链表中是否有环,求环的起点(即链长)和环长. 可以先这么做:首先从起点S出发,给定两个指针,一个快指针一个慢指针,然后每次快指针走1步,慢指针走2步,直到相遇为止.如果已经到达终点/达到规定步数时仍然没有相遇…

 2017 JUST Programming Contest 2.0 题目链接:Codeforces gym 101343 J.Husam and the Broken Present 2 J. Husam and the Broken Present 2 time limit per test:1.0 s memory limit per test:256 MB input:standard input output:standard output After you helped Husam…