杭电oj1032

题目：

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.

给出一个输入n，可以依照上面的算法给出n可以写出的所有数字，总共可以写出的个数叫做“cycle length”，比如22的周期数是16，也就是依照上面的算法，22可以引申出16个数字

大意是判断i，j之间  "the maximum cycle"最大周期数，
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

 

看懂题目很重要，题目不难

1.i，j的大小不确定，需要判断一个大小

2.周期数包括了n本身，所以计数是从1开始

#include 
using namespace std;
int f(int i)
{
    int num=1;//num=0
    if(i==1)
    {
        return 1;
    }
    while(i!=1)
    {
        if(i%2==1)
        {
            i=3*i+1;
        }
        else
        {
            i=i/2;
        }
        num++;
    }
    return num;
    /*else if(i%2==1)
    {
        i=(3*i-1);
    }
    else if(i%2==0)
    {
        i=(i/2);
    }
    return i;*/这个做法会超限
}
int main()
{
    int a,b;
    int a1,b1;
    int n=0;
    int max;//int max=0
    while(cin>>a>>b)
    {
        max=0;
        if(a>b)
        {
            a1=b;
            b1=a;
        }
        else
        {
            a1=a;
            b1=b;
        }
        for(int i=a1;i<=b1;i++)
        {
            n=0;
            /*while(f(i)!=1)
            {
                n++;
            }*/在主函数中给n计数，超限
            if(max