HDU1032——The 3n + 1 problem

The 3n + 1 problem

摇一摇

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34918    Accepted Submission(s): 12648

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
 
 

  

 
 

  题意：
 
 

     这个题意有点难懂，大概是这个样子的:给一个n，如果是1就退出；如果是偶数n=n/2：如果是奇数n=3*n+1；
 
 

  一直循环到n=1。n变的次数为n的周期，初始化是1。现给一个范围i到j，输出i到j内最长的周期。自己写写吧，如果一直WA就看下面的宝典。
 
 

  

 
 

  解题思路：
 
 

     这道题坑点很多：
  1、输入的这两个数不全是前面的小于后面的。
   
  　　2、你为了保证前面的数小于后面的数，可能交换这两个数的值，但是输出的时候要与输入的顺序保持一致。
   
  　　3、编写程序时要注意你设定的步骤变量，最大步骤变量等初始化的位置，以及这些变量比较，累加的位置，当程序层次较多时容易搞迷糊。
   
  　　　　
  1和2如果没注意的话，提交时会让你WA到吐血！
 
 

  

 
 

  

 
 

  代码实现：
 
 

  

 
 

  
  
#include <iostream>
#include <cstdio>
#include <cmath>
#include <queue>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long LL;
const int maxn = 1000000;

int main()
{
    int n, m;
    while( ~scanf("%d%d",&n,&m))
    {
        if( n >  m )
        {
            int t = n; n=m; m=t;
            printf("%d %d ",m,n);
        }
        else printf("%d %d ",n,m);
        LL ans=1, sum=0;
        for( int i=n; i<=m; i++ )
        {
            LL num=i;
            ans = 1;
            while( num!=1 )
            {
                //cout << num << " " ;
                if( num%2==0 )
                    num = num/2;
                else
                    num = 3*num+1;
                ans++;
            }
            sum = max(ans,sum);
        }
        printf("%d\n",sum);
    }
    return 0;
}