100 - The 3n + 1 problem

 The 3n + 1 problem 

Background

Problems in Computer Science are often classified as belonging to acertain class of problems (e.g., NP, Unsolvable, Recursive). In thisproblem you will be analyzing a property of an algorithm whoseclassification is not known for all possible inputs.

The Problem

Consider the following algorithm:

 1.  input n
2.  print n
3.  if n = 1 then STOP
4.   if n is odd then   
5.   else   
6.  GOTO 2

Given the input 22, the following sequence of numbers will be printed22 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 isprinted) for any integralinput value. Despite the simplicity of the algorithm,it is unknown whether this conjecture is true. It has been verified,however, for all integersn such that 0 < n < 1,000,000 (and, in fact,for many more numbers than this.)

Given an input n, it is possible to determinethe number of numbers printed (including the 1). For a givenn this iscalled thecycle-length ofn. In the example above, the cyclelength of 22 is 16.

For any two numbers i and j you are to determine the maximum cyclelength over all numbers betweeni andj.

The Input

The input will consist of a series of pairs of integers i and j, one pair ofintegers per line. All integers will be less than 1,000,000 and greaterthan 0.

You should process all pairs of integers and for eachpair determine the maximum cycle length over all integers between andincludingi andj.

You can assume that no operation overflows a 32-bit integer.

The Output

For each pair of input integers i and j you should output i, j,and the maximum cycle length for integers between and includingi andj. These three numbersshould be separated by at least one space with all three numbers on oneline and with one line of output for each line of input. The integersi andj must appear in the output in the same order in which theyappeared in the input and should befollowed by the maximum cycle length (on the same line).

Sample Input

1 10100 200201 210900 1000

Sample Output

1 10 20100 200 125201 210 89900 1000 174

#include<stdio.h>int test(int n){int t=1;while(n!=1){if(n%2){n=3*n+1;t++;}else{n/=2;t++;}}return t;}int main(void){int i,j,n;while(scanf("%d%d",&i,&j)==2){int max=0;if(i<j){for(n=i;n<=j;n++){if(test(n)>max)max=test(n);}}if(j<i){for(n=j;n<=i;n++){if(test(n)>max)max=test(n);}}if(j==i){max=test(i);}printf("%d %d %d\n",i,j,max);}return 0;}