The 3n + 1 problem

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 32   Accepted Submission(s) : 15
Problem Description
Problems in Computer Science are often classified as belonging to a certain class of problems (e.g., NP, Unsolvable, Recursive). In this problem you will be analyzing a property of an algorithm whose classification is not known for all possible inputs. 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 (including the 1). 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
The input will consist of a series of pairs of integers i and j, one pair of integers per line. All integers will be less than 1,000,000 and greater than 0. 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. You can assume that no opperation overflows a 32-bit integer.
 
Output
For each pair of input integers i and j you should output i, j, and the maximum cycle length for integers between and including i and j. These three numbers should be separated by at least one space with all three numbers on one line and with one line of output for each line of input. The integers i and j must appear in the output in the same order in which they appeared in the input and should be followed by the maximum cycle length (on the same line).
 
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
UVA
 
 #include <stdio.h>
#include <stdlib.h> int main()
{
long long T,N,i,sign,MAX,I,a,b;
while(scanf("%I64d%I64d",&T,&N)!=EOF)
{
a=(T<N)?T:N;
b=(T>N)?T:N;
for(i=a,MAX=;i<=b;i++)
{
I=i;
sign=;
while(I!=)
{
if(I%==)
{
I=*I+;
}
else
{
I=I/;
}
sign++;
}
if(MAX<sign)
MAX=sign;
}
printf("%I64d %I64d %I64d\n",T,N,MAX);
}
return ;
}

The 3n + 1 problem的更多相关文章

  1. UVa 100 - The 3n + 1 problem(函数循环长度)

    题目来源:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=3&pa ...

  2. 烟大 Contest1024 - 《挑战编程》第一章:入门 Problem A: The 3n + 1 problem(水题)

    Problem A: The 3n + 1 problem Time Limit: 1 Sec  Memory Limit: 64 MBSubmit: 14  Solved: 6[Submit][St ...

  3. The 3n + 1 problem 分类: POJ 2015-06-12 17:50 11人阅读 评论(0) 收藏

    The 3n + 1 problem Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 53927   Accepted: 17 ...

  4. uva----(100)The 3n + 1 problem

     The 3n + 1 problem  Background Problems in Computer Science are often classified as belonging to a ...

  5. 【转】UVa Problem 100 The 3n+1 problem (3n+1 问题)——(离线计算)

    // The 3n+1 problem (3n+1 问题) // PC/UVa IDs: 110101/100, Popularity: A, Success rate: low Level: 1 / ...

  6. 100-The 3n + 1 problem

    本文档下载 题目: http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_pro ...

  7. PC/UVa 题号: 110101/100 The 3n+1 problem (3n+1 问题)

     The 3n + 1 problem  Background Problems in Computer Science are often classified as belonging to a ...

  8. UVA 100 - The 3n+1 problem (3n+1 问题)

    100 - The 3n+1 problem (3n+1 问题) /* * 100 - The 3n+1 problem (3n+1 问题) * 作者 仪冰 * QQ 974817955 * * [问 ...

  9. classnull100 - The 3n + 1 problem

    新手发帖,很多方面都是刚入门,有错误的地方请大家见谅,欢迎批评指正  The 3n + 1 problem  Background Problems in Computer Science are o ...

随机推荐

  1. ES6 之 let和const命令 Symbol Promise对象

    ECMAScript 6入门 ECMAScript 6(以下简称ES6)是JavaScript语言的下一代标准,已经在2015年6月正式发布了. (2016年6月,发布了小幅修订的<ECMASc ...

  2. ASCII码对应表chr(num)

    chr(9) tab空格       chr(10) 换行      chr(13) 回车        Chr(13)&chr(10) 回车换行       chr(32) 空格符      ...

  3. jQuery(3)——DOM操作

    ---恢复内容开始---   jQuery中的DOM操作 [DOM操作分类] DOM操作分为DOM Core(核心).HTML-DOM和CSS-DOM三个方面. DOM Core:任何一种支持DOM的 ...

  4. webAppRootKey

    web.xml中webAppRootKey ------------------------------------------------------------------------------ ...

  5. Linux 相关的error处理

    1  dpkg: error: duplicate file trigger interest for filename Notice the first and last lines of /var ...

  6. Design Pattern——单一职责原理

    在类的职责分离上多考虑,做到单一职责,这样的代码才能做到易于维护,易扩展,灵活多样.

  7. Ubuntu 14.0 升级内核到指定版本

    1.卸载现有内核sudo apt purge linux-headers-* linux-headers-*-generic linux-image-*-generic linux-image-ext ...

  8. 下载、安装jdk8(Windows下)并配置变量环境

    一.官网下载地址:http://www.oracle.com/technetwork/java/javase/downloads/index-jsp-138363.html 点击下图中的downloa ...

  9. [妙味JS基础]第七课:运算符、流程控制

    知识点总结 &&(与).||(或).!(非) 与: alert(20 && 20>100) => false alert(20 && 20& ...

  10. Java Lambda表达式入门[转]

    原文链接: Start Using Java Lambda Expressions http://blog.csdn.net/renfufei/article/details/24600507 下载示 ...