The Problem Link.

Below is the description of the problem.

__The 3n + 1 problem__

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:

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

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 before and including

the 1 is printed. 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 and including both 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 10,000 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 operation 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**

100 200

201 210

900 1000

** Sample Output**

100 200 125

201 210 89

900 1000 174

__Approach__

**To solve this Problem, We need to observe two of the following things** -->

#include <iostream>

using namespace std;

#define ll long long

// utility to swap to number

void swap(ll &a, ll &b) {

ll temp = a;

a = b;

b = temp;

}

// utility to find max of two numbers

ll max(ll a, ll b) {

if (a > b)

return a;

return b;

}

int main() {

ll lower_bound, upper_bound;

while (cin >> lower_bound >> upper_bound) {

/*

* Storing the values of our input integers so, that at printing

* Stage we can print the values in the same order as the input values.

*/

ll result_lower_bound = lower_bound, result_upper_bound = upper_bound;

// Checking the condition of our 2nd observation.

if (lower_bound > upper_bound) {

swap(lower_bound, upper_bound);

}

// Finding the maximum path length

ll max_path_length = 0;

for (ll i = lower_bound; i <= upper_bound; i++) {

ll n = i; // Storing i in a temporary variable n

ll current_path_length = 0; // Store the value of current path length

// Loop until the n is not equal to 1

while (n != 1) {

// If n is divisible by 2

if (n % 2 == 0)

n /= 2; // Divide n by 2

else

n = 3 * n + 1; // Change n to 3*n + 1

current_path_length++; // Increase the current Path length by 1

}

max_path_length = max(current_path_length,

max_path_length); // Storing maximum path length in max_path_length

}

// Final stage --> printing result

cout << result_lower_bound << " " << result_upper_bound << " " << max_path_length + 1 << endl;

// max_path_length + 1, because we also include the step when n has become 1.

}

}

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