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 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 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

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 10,000 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.

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).
二、题解

       这题真正的核心部分其实非常水，但是他的输入数据和输出要求有陷阱。首先，输入的时候要计较i和j的大小，然后不要忘了输出的时候也要换过来。这题除了暴力解决以外，还可以用到记忆化存储方法打表。这里有不水的解法http://blog.csdn.net/xieshimao/article/details/6774759。

import java.util.Scanner; 

public class Main { 
	public static int getCycles(int m){
		int sum=0;
		 while(m!=1){
  		   if(m % 2==0){
  			   m=m / 2;
  			   sum++;
  		   }else{
  			   m=3*m+1;
  			   sum++;
  		   }
  	   }
		 return sum+1;
	}
    public static void main(String[] args) { 
       Scanner cin = new Scanner(System.in);
       int s,e,i;
       while(cin.hasNext()){
    	   s=cin.nextInt();
    	   e=cin.nextInt();
    	   boolean flag = false;
    	   if(s > e){
    		   int t=s;
    		   s=e;
    		   e=t;
    		   flag=true;
    	   }
    	   int max=Integer.MIN_VALUE;
    	   for(i=e;i>=s;i--){
    		  int sum=getCycles(i);
	    	  if(sum>max)
    			   max=sum;
    	   }
    	   if(flag){
    		   System.out.println(e+" "+s+" "+max);
    	   }else
    		   System.out.println(s+" "+e+" "+max);
        }
    } 
  } 

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