HDOJ 1032 The 3n+1 problem

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神探特斯拉

最新推荐文章于 2024-02-08 13:38:17 发布

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本文链接：https://blog.youkuaiyun.com/u013555159/article/details/50894787

本文探讨了一个简单算法的循环序列长度问题，通过输入整数n，计算从n开始到1终止的序列中所有数字的打印次数。该文详细介绍了算法流程，包括偶数与奇数情况下的迭代规则，并提供了求解任意区间内最大循环长度的方法。

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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.

//
//  main.cpp
//  3n+1
//
//  Created by 张嘉韬 on 16/3/15.
//  Copyright © 2016年 张嘉韬. All rights reserved.
//

#include <iostream>
#include <cstring>
using namespace std;
int main(int argc, const char * argv[]) {
    int i,j;
    while(scanf("%d%d",&i,&j)!=EOF)
    {
        int s,e,max=0;
        if(i<=j) s=i,e=j;
        else s=j,e=i;
        for(int k=s;k<=e;k++)
        {
            int temp,counter;
            counter=0;
            temp=k;
            //cout<<k<<endl;
            while(temp!=1)
            {
                if(temp%2==0) temp=temp/2;
                else temp=3*temp+1;
                counter++;
                //cout<<"shit"<<endl;
            }
            if(counter>max) max=counter;
        }
        cout<<i<<" "<<j<<" "<<max+1<<endl;
    }
    return 0;
}