Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 207    Accepted Submission(s): 63 Problem Description Problems that process input and generate a simple ``yes'' or ``no'' answer are called decis…

floyd判环算法(龟兔赛跑算法) 注意,这个算法是用来判断一条链+一条环的图,环的长度或者环与链的交界处的,所以此floyd非彼floyd(虽然都是一个人想出来的). (图不是我的) 如果只要求环的长度的话,只要让h和t相遇,然后再让h跑一圈,同时计算出步数就行了. 如果要算出链和环的交界点呢?首先,指针h和t同时从S出发,速度一个为2,一个为1(不要在意细节).当t走到链和环的交界点时,在右边的ht的距离等于st的距离.设st的距离为x,在左边的ht距离为y,那么环的长度就是x+y.现在让h…

Communication 题目链接(点击) 题目描述 The Ministry of Communication has an extremely wonderful message system, designed by the President himself. For maximum efficiency, each office in the Ministry can send a message directly to some, but not necessarily all,…

题目链接:http://acm.sgu.ru/problem.php?contest=0&problem=455 Due to the slow 'mod' and 'div' operations with int64 type, all Delphi solutions for the problem 455 (Sequence analysis) run much slower than the same code written in C++ or Java. We do not gua…

CALCULATOR CONUNDRUM   Alice got a hold of an old calculator that can display n digits. She was bored enough to come up with the following time waster. She enters a number k then repeatedly squares it until the result overflows. When the result overf…

CALCULATOR CONUNDRUM Alice got a hold of an old calculator that can display n digits. She was bored enough to come up with the following time waster. She enters a number k then repeatedly squares it until the result overflows. When the result overflo…

Write an algorithm to determine if a number is "happy". 写出一个算法确定一个数是不是快乐数. A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat th…

Floyd判圈算法 leetcode 上 编号为202 的happy number 问题,有点意思.happy number 的定义为: A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process u…

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…

Floyd判断环算法 全名Floyd’s cycle detection Algorithm, 又叫龟兔赛跑算法(Floyd's Tortoise and Hare),常用于链表.数组转化成链表的题目中. 情景介绍 我们将设置两个指针:slow和fast.slow一次走一格,fast一次走两格. p :环之前的距离 m:S和Q之间的步数 A:链表起点 S:循环起点 Q:初次相遇点 L :环的长度 k :环数 判断是否有环 若在某一时刻slow和fast相遇,则存在环(又可叫Two Pointer…