The 3n + 1 problem 

Background

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.

The Problem

Consider the following algorithm:


1. input n

2. print n

3. if n = 1 then STOP

4. if n is odd then

5. else

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.

The 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 operation overflows a 32-bit integer.

The 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

给出一组数据i,j问在[i,j]中循环长度最大值;
避免重复运算,采用一个数组避免重复...
代码:
 #include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
const int maxn=;
bool vis[maxn];
int s,t;
int main()
{
int cnt,ans,n;
bool flag=;
while(scanf("%d%d",&s,&t)!=EOF){
flag=;
if(s>t){
s^=t;
t^=s;
s^=t;
flag=;
}
memset(vis+s,,sizeof(bool)*(t-s+));
ans=-;
for(int i=s;i<=t;i++)
{
if(!vis[i]){
cnt=;
n=i;
while(n!=){
if(n&) n=*n+;
else n>>=;
cnt++;
if(n>=s&&n<=t) vis[n]=;
}
if(ans<cnt)ans=cnt;
}
}
if(flag)printf("%d %d %d\n",t,s,ans);
else printf("%d %d %d\n",s,t,ans);
}
return ;
}

uva----(100)The 3n + 1 problem的更多相关文章

  1. 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 ...

  2. 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 ...

  3. uva 100 The 3n + 1 problem (RMQ)

    uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem= ...

  4. 【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 / ...

  5. 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 ...

  6. UVa Problem 100 The 3n+1 problem (3n+1 问题)

    参考:https://blog.csdn.net/metaphysis/article/details/6431937 #include <iostream> #include <c ...

  7. UVA 100 The 3*n+1 problem

      UVA 100 The 3*n+1 problem. 解题思路:对给定的边界m,n(m<n&&0<m,n<1 000 000);求X(m-1<X<n+ ...

  8. 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 ...

  9. OpenJudge/Poj 1207 The 3n + 1 problem

    1.链接地址: http://bailian.openjudge.cn/practice/1207/ http://poj.org/problem?id=1207 2.题目: 总时间限制: 1000m ...

  10. The 3n + 1 problem

    The 3n + 1 problem Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) ...

随机推荐

  1. js对数组排序

    <!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8&quo ...

  2. Cheatsheet: 2015 03.01 ~ 03.31

    Web The Architecture of Algolia's Distributed Search Network No promises: asynchronous JavaScript wi ...

  3. git -C

    https://git-scm.com/docs/git -C <path> Run as if git was started in <path> instead of th ...

  4. django 简单的邮件系统

    django邮件系统 Django发送邮件官方中文文档 总结如下: 1.首先这份文档看三两遍是不行的,很多东西再看一遍就通顺了. 2.send_mail().send_mass_mail()都是对Em ...

  5. jquery初涉,First Blood

    jquery可以帮助干的事情有: 遍历HTML文档 操作DOM 处理事件 执行动画 开发Ajax操作 优点就不在这儿扯蛋了~ 1.jquery环境配置 jquery不需要安装,只需要将下载的jquer ...

  6. WPF基础学习第二天(高级控件)

    1.Menu菜单控件 Exp1: Code: <Window x:Class="菜单Menu.MainWindow" xmlns="http://schemas.m ...

  7. Python基础学习笔记(九)常用数据类型转换函数

    参考资料: 1. <Python基础教程> 2. http://www.runoob.com/python/python-variable-types.html 3. http://www ...

  8. Ubuntu 通过Deb安装 MySQL5.5(转载)

    1. 下载 MySQL 5.5 deb 安装包 cd /usr/local/src sudo wget -O mysql-5.5.22-debian6.0-i686.deb http://dev.my ...

  9. 2013 Multi-University Training Contest 4

    HDU-4632 Palindrome subsequence 题意:给定一个字符串,长度最长为1000,问该串有多少个回文子串. 分析:设dp[i][j]表示从 i 到 j 有多少个回文子串,则有动 ...

  10. Java编程思想学习笔记_1(Java内存和垃圾回收)

    1.Java中对象的存储数据的地方: 共有五个不同的地方可以存储数据. 1)寄存器.最快,因为位于处理器的内部,寄存器按需求分配,不能直接控制. 2)堆栈.位于通用RAM,通过堆栈指针可以从处理器那里 ...