110502.c

/*
    110502 - Reverse and Add
    ------------------------

    POPIS

    The reverse and add function starts with a number, reverses its digits, and adds the
    reverse to the original. If the sum is not a palindrome (meaning it does not give the
    same number read from left to right and right to left), we repeat this procedure until
    it does.
    For example, if we start with 195 as the initial number, we get 9,339 as the resulting
    palindrome after the fourth addition:
        195    786   1,473   5,214
        591    687   3,741   4,125
      + —–    + ——    + ——    + ——
        786  1,473   5,214   9,339
    This method leads to palindromes in a few steps for almost all of the integers. But
    there are interesting exceptions. 196 is the first number for which no palindrome has
    been found. It has never been proven, however, that no such palindrome exists.
    You must write a program that takes a given number and gives the resulting
    palindrome (if one exists) and the number of iterations/additions it took to find it.
    You may assume that all the numbers used as test data will terminate in an answer
    with less than 1,000 iterations (additions), and yield a palindrome that is not greater
    than 4,294,967,295.

    INPUT

    The first line will contain an integer N (0 < N ≤ 100), giving the number of test cases,
    while the next N lines each contain a single integer P whose palindrome you are to
    compute.

    SAMPLE INPUT

    3
    195
    265
    750

    OUTPUT

    For each of the N integers, print a line giving the minimum number of iterations to find
    the palindrome, a single space, and then the resulting palindrome itself.

    SAMPLE OUTPUT

    4 9339
    5 45254
    3 6666

    VYSVETLENIE

    Nacita pocet vstupov. Potom po riadkoch cita samotny vstup. Vezmeme nacitane cislo na vstupe 195.
    Z tohto cisla sa spravi reverznute cislo. Tieto dve cisla sa scitaju. Inkrementuje sa pocitadlo o 1,
    aby sme vedeli po kolko scitani sme nasli palindrom.
    Ak vysledne cislo je palindrom (citatelne rovnako zlava-doprava aj zprava-dolava), skoncili sme.
    V opacnom pripade vysledne cislo po sucte vezmeme, spravime jeho reverz a scitame s reverznutym cislom.
    Skontorlujeme ci palindrom, inak opakujeme postup. V pripade cisla 195 sa nasiel po 4 scitani a
    vypise sa pocet scitani a vysledny palindrom: 4 9339*
*/

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

//Princip reverzu
//na vstupe cislo 195
//r = 0 * 10 + 195 % 10 = 5
//num = 195 / 10 = 19
//------repeat--------
//na vstupe cislo 19
//r = 5 * 10 + 19 % 10 = 59
//num = 19 / 10 = 1
//------repeat--------
//na vstupe cislo 1
//r = 59 * 10 + 1 % 10 = 591
//num = 1 / 10 = 0
//Koniec cyklu kvoli podmienke while(num=0)

unsigned long reverse(unsigned long num) {
    unsigned long r=0;
    while(num) {
        r = r * 10 + num % 10;
        num = num/10;
    }
    return r;
}

int main() {
    unsigned long cisel, i,c,c2,j;

    scanf("%lu", &cisel);
    for (i=0; i<cisel; i++) {
        scanf("%lu",&c); //Skenujem konkretne cislo napr. 195

        for(j=1; j<=1000; j++) {//Cyklus bude obmedzeny cislom 1000

            c2=reverse(c); //reverzia cisla 195 cize. 591
            c+=c2; //scitanie do dalsej premennej aj s reverzom cize v bude c= 195 + 591

            c2=reverse(c); //reverz scitaneho aby sa vedelo, ci vysledok v premennej c je naozaj palindrom, cize v c=786 a v c2=687

            if (c==c2) { //porovnanie v pripade zhody cize cislo bude palindrom
                printf("%lu %lu\n", j, c); //vypis a nasledovny break
                break;
            }
        }
    }
    return 0;
}