A solution for the "Reverse and Add" problem

jmftrindade (Joana M. F. da Trindade) — Mon, 14 Apr 2008 01:38:45 GMT

A solution for the "Reverse and Add" problem.

Problem description:
http://icpcres.ecs.baylor.edu/onlinejudge/external/100/10018.html

Author: Joana Matos Fonseca da Trindade
Date: 2008.04.14



/*

 * Solution for the "Reverse and Add" problem.

 * UVa ID: 10018

 */

#include 



#define NDIGITS 100



using namespace std;



/* returns the reversed number */

unsigned long long reverse(unsigned long long number) {

	unsigned long long m = 0; /* reversed number */

	int digits[NDIGITS]; /* digits array */

	int pos = 0, power = 1;

	

	/* init */

	for (int i=0; i		digits[i] = 0;

	}

	

	/* retrieve all digits */

	while (number > 0) {

		digits[pos++] = number % 10;

		number = number / 10;

	}



	/* multiply the reversed digits by the powers of ten */

	for (int i=pos-1; i>=0; i--) {

		m += power * digits[i];

		power *= 10;

	}

	

	return m;

}



/* main */

int main() {

	int nc; /* number of cases */

	int m; /* minimum number of iterations */

	unsigned long long n; /* number */

	

	cin >> nc;

	

	/* reverse and add.. */

	for (int i=0; i		cin >> n;

		m = 0;

		while (true) {

			if (reverse(n) == n) {

				break;

			} else {

				n += reverse(n);

				m++;

			}

		}

		cout << m << " " << n << endl;

	}

	

	return 0;

}

DZone Snippets: 10018 code

A solution for the "Reverse and Add" problem