poj1207
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 50513 | Accepted: 15986 |
Description
Consider the following algorithm:
1. input n 2. print n 3. if n = 1 then STOP 4. if n is odd then n <-- 3n+1 5. else n <-- n/2 6. GOTO 2
Given the input 22, the following sequence of numbers will be printed 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1
It is conjectured that the algorithm above will terminate (when a 1 is printed) for any integral input value. Despite the simplicity of the algorithm, it is unknown whether this conjecture is true. It has been verified, however, for all integers n such that 0 < n < 1,000,000 (and, in fact, for many more numbers than this.)
Given an input n, it is possible to determine the number of numbers printed before the 1 is printed. For a given n this is called the cycle-length of n. In the example above, the cycle length of 22 is 16.
For any two numbers i and j you are to determine the maximum cycle length over all numbers between i and j.
Input
You should process all pairs of integers and for each pair determine the maximum cycle length over all integers between and including i and j.
Output
Sample Input
1 10
100 200
201 210
900 1000
Sample Output
1 10 20
100 200 125
201 210 89
900 1000 174
Source
poj1207的更多相关文章
- poj-1207 THE 3n+1 problem
Description Problems in Computer Science are often classified as belonging to a certain class of pro ...
- poj1207 3n+1 problem
The 3n + 1 problem Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 60496 Accepted: 19 ...
- Poj1207 The 3n + 1 problem(水题(数据)+陷阱)
一.Description Problems in Computer Science are often classified as belonging to a certain class of p ...
- ACM训练计划建议(写给本校acmer,欢迎围观和指正)
ACM训练计划建议 From:freecode# Date:2015/5/20 前言: 老师要我们整理一份训练计划给下一届的学弟学妹们,整理出来了,费了不少笔墨,就也将它放到博客园上供大家参考. 菜 ...
- 【POJ水题完成表】
题目 完成情况 poj1000:A+B problem 完成 poj1002:电话上按键对应着数字.现在给n个电话,求排序.相同的归一类 完成 poj1003:求最小的n让1+1/2+1/3+...+ ...
- poj 算法 分类
转载请注明出处:優YoU http://blog.csdn.net/lyy289065406/article/details/6642573 最近AC题:2528 更新时间:2011.09.22 ...
- POJ1008 1013 1207 2105 2499(全部水题)
做了一天水题,挑几个还算凑合的发上来. POJ1008 Maya Calendar 分析: #include <iostream> #include <cstdio> #inc ...
- POJ 水题(刷题)进阶
转载请注明出处:優YoU http://blog.csdn.net/lyy289065406/article/details/6642573 部分解题报告添加新内容,除了原有的"大致题意&q ...
- ACM训练计划建议(转)
ACM训练计划建议 From:freecode# Date:2015/5/20 前言: 老师要我们整理一份训练计划给下一届的学弟学妹们,整理出来了,费了不少笔墨,就也将它放到博客园上供大家参考. 菜 ...
随机推荐
- Java Thread 总结
目 录 线程的概述(Introduction) 线程的定义(Defining) 1) 继承java.lang.Thread类 2) 实现java.lang.Runnable接口 线程的启动(St ...
- 反调试技术(Delphi版)
1.程序窗口句柄检测原理:用FindWindow函数查找具有相同窗口类名和标题的窗口,如果找到就说明有OD在运行//****************************************** ...
- EasyUI Combotree 只允许选择 叶子节点
$("#SDID").combotree({ url: '/Ajax/GetDeptTree.aspx?level=4&pid=-1', onSelect: functio ...
- Node.js log3:create ejs engine and jade engine
The base condition is ensure that you have installed express. 1.create ejs engine Using windows d ...
- cdoj 491 Tricks in Bits
//无脑爆居然能过!!!!! 解:其实正解也是暴力,但是可以证明在n>6时答案一定为零. 第一步:对于任意两个数他们的二进制数要么有一半+的位是相同的,要么有一半+的位是不同的,于是首先使用与运 ...
- uva10820 send a table (nlogn求1-n欧拉函数值模版
//重点就是求1-n的欧拉函数啦,重点是nlogn求法的版 //大概过程类似于筛选法求素数 #include<cstdio> #include<iostream> #inclu ...
- OpenstackUbuntu
1,create user
- hdu 1978 How many ways(dp)
Problem Description 这是一个简单的生存游戏,你控制一个机器人从一个棋盘的起始点(1,1)走到棋盘的终点(n,m).游戏的规则描述如下: 1.机器人一开始在棋盘的起始点并有起始点所标 ...
- poj 3255 求次大最短路
Roadblocks Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 5508 Accepted: 2088 Descri ...
- HTML5新特性学习-2
本文在于巩固基础 HTML5绘图基础 <canvas>画布元素的使用 <div> <canvas id="can" width="200px ...