UVA 100 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

题意：输入两个整数，然后输出这两个数，这两个数和他们之间的数进行如下操作：这个数是不是偶数，如果是这个数就除以2，否则乘以3加1，直到这个数变为1结束。记录这个数变到1　　　　 　　　　　　　　　　　　     需要几步，包括这个数本身和变为1的那一步，可以参考题目给出的例子。然后输出这两个数及他们之间的数转换为1所需要的步骤的最大值。 分析：这题需要注意的有： 　　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！

AC源代码（C语言）：

 

 1 #include<stdio.h>
 2 
 3 int main()
 4 {
 5     int j,MAX,a,b,i,t;
 6     while(scanf("%d%d",&a,&b)==2)
 7     {
 8 
 9         MAX=0;
10         if(a>b)
11         {
12             printf("%d %d ",a,b);     /*要按输入a和b的顺序输入，否则提交时会WA*/
13             t=a;
14             a=b;
15             b=t;
16         }
17         else printf("%d %d ",a,b);      /*这里的else不能包含下面的语句，因为下面的语句是a大于b或者a小于等于b都需要操作的*/
18         for(i=a;i<=b;i++)
19         {
20             a=i;
21             j=1;                    /*在这里进行j的初始化，j要初始化为1，不能为0*/
22             while(a!=1)
23             {
24                 if(a%2==0)
25                 {
26                     a/=2;
27                 }
28                 else
29                 {
30                     a=3*a+1;
31                 }
32                 j++;
33             }
34             if(j>MAX)
35                 MAX=j;
36         }
37         printf("%d\n",MAX);
38     }
39     return 0;
40 }