[ACM]The 3n + 1 problem

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

 

Source

UVA

解题思路：

每一个数都有一个序列（以1结束），题目要求是输入两个数 i ,j ，输出的是从i到j （包括他们）之间所有数序列的最大长度。外部用一层循环就可以了，根据所给的运算法则对每一个数进行运算，每一次循环都用count来记录当前数的序列长度。

代码：

#include <iostream>
using namespace std;
int main()
{
    int i,j;
    while(cin>>i>>j)
    {
       int tempi,tempj;
       tempi=i;tempj=j;
       int max=0;
       tempi=(i<=j)?i:j;//把i,j中小的数给tempi
       tempj=(i>=j)?i:j;
       for(;tempi<=tempj;tempi++)
       {
           int tempii//定义这个变量很关键，因为每一次循环过程中,tempi的值都不能改变，用tempii来代替tempi，进行运算法则运算，这样不会破坏外层循环的连续性。
           int count;
           count=1;
           tempii=tempi;
           while(tempii!=1)
           {
               if(tempii%2==0)
                {
                    tempii=tempii/2;
                    count++;
                }
                else
                {
                    tempii=3*tempii+1;
                    count++;
                }
           }
           if(max<=count)
            max=count;
       }
       cout<<i<<" "<<j<<" "<<max<<endl;
    }
    return 0;
}

运行结果：