100 – The 3n + 1 problem Solution

This is an easy problem from uva judge http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=36

Brute Force, Iterative, Complete search.

How I did it

• I read i and j.
• I save the values of i and j in oi = i and oj =j. (Just for the output).
• If i is greater than j swap i and j.
• Traverse i to j and calculate the cycle length of the current number.
• Save the max length cycle.
• Print oi, oj and the max length cycle.
• Accepted! 😀

Hints

• j can be greater than i.
• Print i and j in the original order. Print j i will give us wrong answer.

Code

#include <bits/stdc++.h>

using namespace std;

int  getCycle(int n)
{
    int cont = 1;

    while(n != 1)
    {
        if(n%2 != 0)
          n = n*3 + 1;
        else
            n/=2;
        cont++;
    }
return cont;
}

int main()
{
    int i,j, cycleL, ans;
    int io, jo;

    while(scanf("%d %d", &i, &j) != EOF)
    {
        io = i, jo = j;

        if(i > j)
           swap(i,j);

        ans = -1000;
        for(int x=i; x<=j; x++)
        {
            cycleL = getCycle(x);
            ans = max(cycleL, ans);
        }

        printf("%d %d %d\n", io, jo, ans);
    }
return 0;
}