A - The 3n + 1 problem （HDU 1032）

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 operation 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  

题解

对题目的理解：

1. 循环长度：对于某一个正整数n，进行3n+1（n为奇数）或者2n（n为偶数）的操作，直到得到1，操作的次数就是循环长度
2. 最大循环长度：计算在所给的m和n之间（包括m，n）的整数的循环长度，比较这些数的循环长度，得出最大值

思路：

• 确定m，n谁大谁小。遍历其间的每一个数，并进行如下操作n1 = (n1%2 == 1) ? 3*n1+1 : n1/2;直到得到1，同时记录操作的次数。将每一次得到的循环次数与之前的最大次数比较，更新目前的最大循环长度。

Code

    #include<iostream>

	using namespace std;

	int main()
	{
	    int tem , m , n , min, max, cnt;
	    long long n1;
    	while(cin>> n >> m)
    	{
    	    max = m > n ? m : n;
    	    min = m < n ? m : n;
    	    tem = 0;
    	    for(int i = min; i <= max; i++)
    	    {
    	        n1 = i;
    	        cnt = 1;
    	        while(n1 != 1)
    	        {
    	            n1 = (n1%2 == 1) ? 3*n1+1 : n1/2;
    	            cnt++;
    	        }
    	        tem = tem > cnt ? tem : cnt;     
    	    }
    	    cout<<n<<" "<<m<<" "<<tem<<endl;
    	}
    	return 0;
	}