The Necklace Problem

demolaynyc

Nov 21 2006, 06:31 PM

Hey, I need help! Can someone code this problem in C++

The instructions are:

An interesting problem in number theory is sometimes called the "necklace problem." This problem begins with two single-digit numbers. The next number is obtained by adding the first two numbers together and saving only the ones-digit. This process is repeated until the "necklace" closes by returning to the original two numbers. For example, if the starting numbers are 1 and 8, twelve steps are required to close the "necklace":

1 8 9 7 6 3 9 2 1 3 4 7 1 8

Here's another program:

It says to...

modify the program to determine what integers of two, three, or four digits are equal to the sum of the cubes of their digits

Here's the code:

CODE

/* Chapter 4 Exercise 9 by Albert Villaroman 11-20-06 */

#include <iostream.h>

int main() {

    long numValue; //user input
    cout <<"Enter an integer: ";
      cin >>numValue;

    long MaxDigits = 100000000;

    while (MaxDigits > numValue) { //while loop to get how many digits numValue has
        MaxDigits /= 10;
    }

    long sum = 0;    //sum of all digits
    while (MaxDigits >=1) {
        long digit = numValue / MaxDigits; //get digit
        sum += digit*digit*digit;
        numValue = numValue % MaxDigits; //enter new value for numValue
        MaxDigits /= 10;       //ex: 100 => 10
    }



    cout <<"Sum of the cubes of the digits is " <<sum <<endl;
    return(0);
}

pyost

Nov 21 2006, 07:15 PM

Since I have never worked in C++, forgive me if there are any errors, since I'm going through a tutorial in order to solve this.

CODE

#include <iostream>

int main ()
{
  // Variables
  short first;
  short second;
  short previous;
  short current;
  short help;

  // Input and communication with the user
  cout << "Insert the first digit: ";
  cin >> first;
  cout << "\n";
  cout << "Insert the second digit: ";
  cin >> second;
  cout << "\n";
  cout << "The necklace is: ";

  // Preparing for the loop
  previous = second;
  current = (first + second)%10;
  cout >> first;
  cout >> second;

  // And not the WHILE loop

  while ( (previous != first) || (current != second) )
    {
      cout >> current;
      help = previous;
      previous = current;
      current = (help + current)%10;
    }

  // Writing the last digit
  cout >> current;

  return 0;
}

So, what are we doing here? First, the user inputs two digits. Next, we create what the second and the third digit would be. After that, it's all in the loop. We create the next number if we haven't reached the first two. When we have, the program won't enter the loop, and will write the last remaining digit (which equals the second digit).

demolaynyc

Nov 22 2006, 06:29 PM

wow. no experience in c++ yet you have the answer. such skill. Thanks a lot man, ill go try that out.

pyost

Nov 22 2006, 06:49 PM

It's not all about knowing the language, but knowing how to solve the problem

That's what I've been doing in my school for the past year. We are using Pascal, but doesn't stop us from solving extremely hard problems

The necklace one is easy

ADD REPLY / Got an Opinion!	a humble request :-)	RAPID SEARCH!	Free Hosting	[X]
Express your Opinions, Thoughts or Contribute more info. to help others. Ask your Doubts & Queries to get answers, So that "Together We can help others!"	Register FREE for AD-FREE forum, Create your own topics, Ask Questions, track topics, setup subscriptions & notifications and Get a Free Website w/ Email and FTP.		500MB Space No Ads, CPanel, FTP, PHP, MySQL, EMails - 100% FREE