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Program not working, not giving output, I don't know what to do, where the problem is.
I'm trying to find out the largest palindrome made from the product of two 3-digit numbers.
#include <stdio.h>
main() {
int i, k, j, x;
long int a[1000000], palindrome[1000000], great, sum = 0;
// for multiples of two 3 digit numbers
for (k = 0, i = 100; i < 1000; i++) {
for (j = 100; j < 1000; j++) {
a[k] = i * j; // multiples output
k++;
}
}
for (i = 0, x = 0; i < 1000000; i++) {
// for reverse considered as sum
for (; a[i] != 0;) {
sum = sum * 10 + a[i] % 10;
}
// for numbers which are palindromes
if (sum == a[i]) {
palindrome[x] = a[i];
x++;
break;
}
}
// comparison of palindrome number for which one is greatest
great = palindrome[0];
for (k = 0; k < 1000000; k++) {
if (great < palindrome[k]) {
great = palindrome[k];
}
}
printf("\ngreatest palindrome of 3 digit multiple is : ", great);
}
What do you mean with "not working"?
There are two things, from my point of view:
1) long int a[1000000], palindrome[1000000]
Depending on you compile configuration you could have problems compiling your code.
Probably the array is too big to fit in your program's stack address space.
In C or C++ local objects are usually allocated on the stack. Don't allocate it local on stack, use some other place instead. This can be achieved by either making the object global or allocating it on the global heap.
#include <stdio.h>
long int a[1000000], palindrome[1000000], great, sum = 0;
main() {
int i, k, j, x;
2) printf("\ngreatest palindrome of 3 digit multiple is : ", great);
I will change it by :
printf("\ngreatest palindrome of 3 digit multiple is %li: ", great);
Regards.
Compiling and running your code on an on-line compiler I got this:
prog.c:3:1: warning: type specifier missing, defaults to 'int' [-Wimplicit-int]
main() {
^
prog.c:34:61: warning: data argument not used by format string [-Wformat-extra-args]
printf("\ngreatest palindrome of 3 digit multiple is : ", great);
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ^
2 warnings generated.
Killed
Both the warnings should be taken into account, but I'd like to point out the last line. The program was taking too much time to run, so the process was killed.
It's a strong suggestion to change the algorithm, or at least to fix the part that checks if a number is a palindrome:
for (; a[i] != 0;) { // <-- If a[i] is not 0, this will never end
sum = sum * 10 + a[i] % 10;
}
I'd use a function like this one
bool is_palindrome(long x)
{
long rev = 0;
for (long i = x; i; i /= 10)
{
rev *= 10;
rev += i % 10;
}
return x == rev;
}
Also, we don't need any array, we could just calculate all the possible products between two 3-digits number using two nested for loops and check if those are palindromes.
Starting from the highest numbers, we can store the product, but only if it's a palindrome and is bigger than any previously one found, and stop the iteration of the inner loop as soon as the candidate become less then the stored maximum. This would save us a lot of iterations.
Implementing this algorithm, I found out a maximum value of 906609.
I'm a bit stuck on one of my problems not because I don't know, but because I can't use more complex operations.(functions and multiple arrays)
So I need to make a program in C that ask for an input of an array(max 100 elements) and then program needs to sort that matrix by numbers with same digits.
So I made everything that I know, I tested my program with sorting algorithm from minimum to maximum values and it works, only thing that I can't understand is how should I test if the number have same digits inside the loop? (I can't use functions.)
So I know the method of finding if the number have the same digits but I don't know how to compare them. Here is an example of what I need.
This is what I have for now this sorts numbers from min to max.
#include <stdio.h>
int main() {
int matrix[100];
int i,j;
int temp,min;
int elements_number=0;
printf("Enter the values of matrix-max 100 elements-type -1 to end: ");
for(i=0;i<100;i++){
scanf("%d",&matrix[i]);
elements_number++;
if(matrix[i]==-1){
elements_number--;
break;
}
}
for (i=0; i<elements_number; i++) {
min=i;
for (j=i+1; j<elements_number; j++) {
if (matrix[j] < matrix[min])
min = j;
}
temp = matrix[i];
matrix[i] = matrix[min];
matrix[min] = temp;
}
for(i=0;i<elements_number;i++){
if(i!=elements_number-1){
printf("%d,",matrix[i]); }
else printf("%d.",matrix[i]);
}
return 0;
}
I need this output for these numbers:
INPUT :
1 22 43 444 51 16 7 8888 90 11 -1
OUTPUT:
1,22,444,7,8888,11,43,51,16,90.
Integers with 1 digit count as "numbers with same number of digits" like 7 and 1 in this example.
Hope that you can help.
After processing the array, the single-digit numbers should all be in the left part of the array, the other numbers in the right part. Within each part, the original order of the elements should be preserved. This is called a stable partition. It is different from sorting, because the elements are only classified into two groups. Sorting means that there is a clear relationship between any two elements in the array.
This can be done by "filtering" the array for single-digit numbers and storing the other numbers that were filtered out in a temporary second array. Then append the contents of that second array to the (now shorter) first array.
Here's how that could work:
#include <stdlib.h>
#include <stdio.h>
void print(const int *arr, int n)
{
for (int i = 0; i < 10; i++) {
if (i) printf(", ");
printf("%d", arr[i]);
}
puts(".");
}
int is_rep_digit(int n)
{
int q = n % 10;
n /= 10;
while (n) {
if (n % 10 != q) return 0;
n /= 10;
}
return 1;
}
int main()
{
int arr[10] = {1, 22, 43, 444, 51, 16, 7, 8888, 90, 11};
int aux[10]; // auxliary array for numbers with several digits
int i, j, k;
print(arr, 10);
j = 0; // number of single-digit numbers
k = 0; // number of other numbers
for (i = 0; i < 10; i++) {
if (is_rep_digit(arr[i])) {
arr[j++] = arr[i]; // pick single-digit number
} else {
aux[k++] = arr[i]; // copy other numbers to aux
}
}
k = 0;
while (j < 10) { // copy aux to end of array
arr[j++] = aux[k++];
}
print(arr, 10);
return 0;
}
Edit: I've just seen your requirement that you can't use functions. You could use Barmar's suggestion to test divisibility by 1, 11, 111 and so on. The tricky part is to find the correct divisor, however.
Anyway, the point I wanted to make here is that you don't need a full sorting algorithm here.
Hello guys i am trying to implement a program which is finding the happy numbers were between two numbers A and B.
Summing the squares of all the digits of the number, we replace the number with the outcome, and repeat the process. If after some steps the result is equal to 1 (and stay there), then we say that the number N is **<happy>**. Conversely, if the process is repeated indefinitely without ever showing the number 1, then we say that the number N is **<sad>**.
For example, the number 7 is happy because the procedure described above leads to the following steps: 7, 49, 97, 130, 10, 1, 1, 1 ... Conversely, the number 42 is sad because the process leads to a infinite sequence 42, 20, 4, 16, 37, 58, 89, 145, 42, 20, 4, 16, 37 ...
I try this right down but i am getting either segm faults or no results.
Thanks in advance.
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
void happy( char * A, int n);
int numPlaces (long n);
int main(void)
{
long A,B;
int npA;
char *Ap;
printf("Give 2 Numbers\n");
scanf("%li %li",&A,&B);
npA = numPlaces(A);
Ap = malloc(npA);
printf("%ld %d\n",A,npA);
//Search for happy numbers from A to B
do{
sprintf(Ap, "%ld", A);
happy(Ap,npA);
A++;
if ( npA < numPlaces(A) )
{
npA++;
Ap = realloc(Ap, npA);
}
}while( A <= B);
}
//Finds happy numbers
void happy( char * A, int n)
{
//Basic Condition
if ( n == 1)
{
if (A[0] == 1 || A[0] == 7)
printf("%c\n",A[0]);
printf("%s\n",A);
return;
}
long sum = 0 ;
char * sumA;
int nsum;
int Ai;
//Sum the squares of the current number
for(int i = 0 ; i < n;i++)
{
Ai = atoi(&A[i]);
sum = sum + (Ai*Ai);
}
nsum = numPlaces (sum);
sumA = malloc(nsum);
sprintf(sumA, "%li", sum);
happy(sumA,nsum);
free(sumA);
}
//Count digits of a number
int numPlaces (long n)
{
if (n < 0) return 0;
if (n < 10) return 1;
return 1 + numPlaces (n / 10);
}
Thanks for your time.
by the definition of your program sad numbers will cause your program to run forever
Conversely, if the process is repeated indefinitely
You need to add a stopping condition, like if I have looped for 1000 times, or if you hit a well known non terminating number (like 4) (is there a definite list of these? I dont know)
I find this solution tested and working..
Thanks for your time and I am sorry for my vagueness.
Every advice about this solution would be welcome
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
void happy( char * A, int n);
int numPlaces (long n);
int happynum = 0;
int main(void)
{
long A,B;
int npA;
char *Ap;
printf("Give 2 Numbers\n");
scanf("%li %li",&A,&B);
npA = numPlaces(A);
Ap = malloc(npA);
//Search for happy numbers from A to B
do{
sprintf(Ap, "%ld", A);
happy(Ap,npA);
if (happynum ==1)
printf("%s\n",Ap);
A++;
if ( npA < numPlaces(A) )
{
npA++;
Ap = realloc(Ap, npA);
}
}while( A <= B);
}
//Finds happy numbers
void happy( char * A, int n)
{
//Basic Condition
if ( n == 1)
{
if (A[0] == '3' || A[0] == '6' || A[0] == '9')
{
happynum = 0;
}
else
{
happynum = 1;
}
return;
}
long sum = 0;
char * sumA;
int nsum;
int Ai;
//Sum the squares of the current number
for(int i = 0 ; i < n;i++)
{
Ai = (int)(A[i]-48);
sum = sum + (Ai*Ai);
}
nsum = numPlaces (sum);
sumA = malloc(nsum);
sprintf(sumA, "%li", sum);
happy(sumA,nsum);
free(sumA);
}
//Count digits of a number
int numPlaces (long n)
{
if (n < 0) return 0;
if (n < 10) return 1;
return 1 + numPlaces (n / 10);
}
Your code uses some questionable practices. Yoe may be misguided because you are concerned about performance and memory usage.
When you allocate memory for the string, you forget to allocate one character for the null terminator. But you shouldn't be allocating, re-allocating and freeing constantly anyway. Dynamic memory allocation is expensive compared to your other operations.
Your limits are long, which may be a 32-bit or 64-bit signed integer, depending on your platform. The maximum number that can be represented with e 64-bit signed integer is 9,223,372,036,854,775,807. This is a number with 19 digits. Add one for the null terminator and one for a possible minus sign, so that overflow won't hurt, you and use a buffer of 21 chars on the stack.
You probably shouldn't be using strings inthe first place. Use the basic code to extract the digits: Split off the digit by taking the remainder of a division by 10. Then divide by 10 until you get zero. (And if you use strings with a fixed buffer size, as described above, you don't have to calculate the difits separately: sprintf returns the number of characters written to the string.
Your functions shouldn't be recursive. A loop is enough. As pm100 has noted, you need a termination criterion: You must keep track of the numbers that you have already visited. Each recursive call creates a new state; it is easier to keep an array, that can be repeatedly looked at in a loop. When you see a number that you have already seen (other than 1, of course), your number is sad.
Happy and sad numbers have this property that when your sum of squares is a number with a known happiness, the original number has this happiness, too. If you visit a known das number, the original number is sad. If you visit a known happy number, the original number is happy.
The limits of your ranges may ba large, but the sum of square digits is not large; it can be at most the number of digits times 81. In particular:
type max. number number of max. square sum dss
int 2,147,483,647 1,999,999,999 730
uint 4,294,967,295 3,999,999,999 738
long 9,223,372,036,854,775,807 8,999,999,999,999,999,999 1522
ulong 18,446,744,073,709,55,1616 9,999,999,999,999,999,999 1539
That means that when you take the sum of digit squares of an unsigned long, you will get a number that is smaller than 1540. Create an array of 1540 entries and mark all known happy numbers with 1. Then you can reduce your problem to taking the sum of digit squares once and then looking up the happiness of the number in this array.
(You can do the precalculation of the array once when you start the program.)
UPDATE: Thanks a lot M Oehm for your awesome answer, really helped me a lot. That struct pos moves is really helpful, haven't studied it yet on class. Im working on a fully solution of my code adding yours as a bone on the programs skeleton. Already fixed the problem of updating fila and columna, the random choice and the switch from 0 to 7 without the ' ' because they're not characters as you and davidc pointed. My program still have some problems which im working on before posting the fully operational program here. I'll update the code tomorrow if not today. Thank you all for your comments and solutions, and M oehm for the time you spent making that wonderful answer.
------------------------------------------------------------------------------------------------------------------------------------
UPDATE 2: Finnished it, made some little changes on M Oehm code and instead of putting manually the first location of the horse i used my previous PosicionCaballo(). Had to delete the code of the MoverCaballo() which had a Switch with 8 possible moves which was set by a random number because i couldn't make it work (i guess the main problem was that part because already was a mess). Now the program with the code below should ask the user for the initial position of the horse, after that the screen will print a 10x10 table filled with 0's (free spaces), fill it with 1's (taken spaces which did the horse moving randomly) and when it finnishes display a message on how many positions did it take.
COORD cxy;
#define posicion(x,y) {(cxy.X)= (x);(cxy.Y)= (y); SetConsoleCursorPosition((GetStdHandle(STD_OUTPUT_HANDLE)), (cxy) );}
int ajedrez[10][10];
int fila, columna;
void GenerarTablero(int m[10][10]){
int i, j;
for (i = 0; i < 10; i++){
for (j = 0; j < 10; j++){
m[i][j] = 0;
}
}
}
GenerarTablero makes the chess board and fills it with 0's.
Global are fila (row) and columna (column) the parts of the ajedrez[10][10]
ajedrez[10][10] is the chess table in a 10x10 size.
void PosicionCaballo(int m[10][10]){
printf("Fila: ");
scanf_s("%d", &fila);
printf("Columna: ");
scanf_s("%d", &columna);
printf("\n");
m[fila][columna] = 2;
system("cls");
}
PosicionCaballo does ask the user for the initial position of the horse and puts the horse on the table.
Example: Row: 5 Column: 5
int horse(int y, int x)
{
int visited[SIZE][SIZE] = { { 0 } };
int count = 0;
if (on_board(y, x) == 0) return -1;
/* Set starting position */
visited[y][x] = 1;
while (1) { /* Infinite loop - must use break */
int poss[8]; /* Possible moves */
int nposs = 0; /* Actual length of poss */
int i, k = 1;
for (i = 0; i < 8; i++) {
int xx = x + moves[i].x;
int yy = y + moves[i].y;
if (on_board(yy, xx) && visited[yy][xx] == 0) {
poss[nposs++] = i;
}
}
/* No more valid moves: return */
if (nposs == 0){
posicion(0, 11);
printf("El tablero se ha bloqueado con ");
return count;
}
/* pick one of the valid moves */
i = poss[rand() % nposs];
x = x + moves[i].x;
y = y + moves[i].y;
/* update horse's position */
visited[y][x] = 1;
count++;
/* print position */
posicion(y, x);
printf("1");
}
return -1; /* Should never be reached */
}
void MostrarMapa(int m[10][10]){
int i, j;
for (i = 0; i < 10; i++){
for (j = 0; j < 10; j++){
printf("%d", ajedrez[i][j]);
}
printf("\n");
}
}
MostrarMapa only prints the chess table on the screen.
int main(void){
int n;
srand(time(NULL));
GenerarTablero(ajedrez);
PosicionCaballo(ajedrez);
MostrarMapa(ajedrez);
n = horse(fila, columna);
posicion(31, 11);
printf("%d movimientos\n", n);
getch();
return 0;
}
and then my main which im using all the functions stated upside.
Thank you very much in advance for your help guys :).
I guess that your assignment is about finding a valid path that visits all squares. Your code tries to find one random path.
Your code has several errors:
When you test ajedrez[fila - 2][columna - 1], you don't check whether fila - 2 or columna - 1 are really valid indices of your chess board. If you access invalid indices, way -1 or 11, you invoke undefined behaviour.
You don't update fila and columna, that is: You don't move your horse.
You overwrite the board twice. That's not an error, but your code shouldn't do double duty.
Your random choice is broken. You have eight possible moves, so you want rand() % 8, which yields numbers from 0 to 7. (David has already pointed this out in a comment.)
Your case labels are charater constants, not numbers. Use case 0:, not case '0':.
You just skip invalid moves. When there are no more valid moves, this will result in an infinite loop. You should check this condition and terminate the loop if it occurs.
As I understand it, the ´posicion` macro is just to show ehere the horse is. Maybe you should skip that at the moment and just print the new coordinates, which isn't as pretty, but straightforward.
Your eight switch cases manifest another flaw: You have the same repeated code eight times over. The only difference is the jump pattern. Such a set-up lends itself to either writing a function where you pass the row and columns distance to jump or to using an array of possible jump patterns.
Your code should instead do something like this for each move:
Loop over all eight jump patterns. If the horse would jump off the board or if the horse would visit a tile that has already been visited, skip that possibility. Otherwise, add the move to an auxiliary array.
If the number of possibilities is zero, terminate the loop - the horse has nowhere to go.
Pick one of the valid moves.
Move the horse, mark the current tile visited.
Report the jump, if desired: Print the new position or place the cursor, whatever.
Below is an example implementation that uses an array of jump patterns. It will give one random path. You can adapt this code to your problem.
#include <stdlib.h>
#include <stdio.h>
#include <time.h> /* for time() */
#define SIZE 10 /* Fixed board size */
struct pos {
int x, y;
};
struct pos moves[8] = { /* Jump patterns */
{1, 2},
{2, 1},
{2, -1},
{1, -2},
{-1, -2},
{-2, -1},
{-2, 1},
{-1, 2}
};
/*
* Is position (y, x) a valid board coordinate?
*/
int on_board(int y, int x)
{
if (x < 0 || x >= SIZE) return 0;
if (y < 0 || y >= SIZE) return 0;
return 1;
}
/*
* Move the horse randomly, starting from (y, x). Print the
* visited fields and return the number of moves made or
* -1 if an error occurs.
*/
int horse(int y, int x)
{
int visited[SIZE][SIZE] = {{0}};
int count = 0;
if (on_board(y, x) == 0) return -1;
/* Set starting position */
visited[y][x] = 1;
printf("%c%d, ", 'A' + y, x + 1);
while (1) { /* Infinite loop - must use break */
int poss[8]; /* Possible moves */
int nposs = 0; /* Actual length of poss */
int i;
for (i = 0; i < 8; i++) {
int xx = x + moves[i].x;
int yy = y + moves[i].y;
if (on_board(yy, xx) && visited[yy][xx] == 0) {
poss[nposs++] = i;
}
}
/* No more valid moves: return */
if (nposs == 0){
printf("whoa!\n");
return count;
}
/* pick one of the valid moves */
i = poss[rand() % nposs];
x = x + moves[i].x;
y = y + moves[i].y;
/* update horse's position */
visited[y][x] = 1;
count++;
/* print position */
printf("%c%d, ", 'A' + y, x + 1);
}
return -1; /* Should never be reached */
}
int main()
{
int n;
srand(time(NULL));
n = horse(3, 6);
printf("%d moves\n", n);
return 0;
}
You seem to update the position using posicion(), but not update fila and columna
So on Project Euler the Problem 4 states the following:
A palindromic number reads the same
both ways. The largest palindrome made
from the product of two 2-digit
numbers is 9009 = 91 99.
Find the largest palindrome made from
the product of two 3-digit numbers.
I have tried the following:
#include <stdio.h>
#include <stdlib.h>
int check(int result)
{
char b[7];
sprintf(b, "%d", result);
if (b[0] == b[5] && b[1] == b[4] && b[2] == b[3])
{
return 1;
}
else
{
return 0;
}
}
int main () {
int i;
int g;
int final;
for (i = 999; i > 99; i--)
{
for (g = 999; g > 99; g--)
{
if (check(g*i) == 1)
{
final = g*i;
goto here;
}
}
}
here:
printf("%d", final);
}
But, this does not work. Instead of the right answer, I get 580085, which I guess is a palindrome at least, but still not the right answer.
Let me explain my program starting from int main:
int i and int g are my multipliers. They are those two three digit numbers.
int final is the number that will store the largest palindrome.
I start two for loops going to down to get every number possibility.
I get out of the loop using a goto when the first palindrome is reached(probably should not but, it doesn't effect a small program like this too much).
The first palindrome should be the biggest one possible since I am counting down from the top.
Let me now explain my check:
First off since these are two three digit numbers multiplying together to determine the size a char would need to be to hold that value I went to a calculator and multiplied 999 * 999 and it ended up being 6 then I need to add one because I found out from one the questions I posted earlier that sprintf puts a \0 character at the end.
Ok, now that I have a char and all, I copied result (which i*g in int main) and put it in char b[7].
Then I just checked b to see if it equalled it self with by hard coding each slot I needed to check for.
Then I returned accordingly, 1 for true, and 2 for false.
This seems perfectly logical to me but, it does not work for some weird reason. Any hints?
This assumption is wrong:
The first palindrome should be the biggest one possible since I am counting down from the top.
You will check 999*100 = 99900 before 998*101 = 100798, so clearly you can´t count on that.
The problem is that the first palindrome that you find is not the bigger one for sure.
Just an example:
i = 900, g = 850 -> 765000
i = 880, g = 960 -> 844800
The first one is smaller, but since you iterate first on i, then on g it will be discovered first.
Ok, they are not palindrome but the concept is the same..
I think you are tackling this problem back to front. It would be more efficient to generate the palindromes from highest to lowest then check by factorizing them. First one that has two three digit factors is the answer.
e.g.
bool found = false;
for (int i = 998; i >= 100; i--)
{
char j[7];
sprintf(j,"%d",i);
j[3]= j[2];
j[4]= j[1];
j[5]= j[0];
int x =atoi(j);
int limit = sqrt((float) x);
for (int z = 999; z >= limit; z--)
{
if (x%z==0){
printf("%d",x);
found = true;
break;
}
}
if (found) break;
}
The first palindrome should be the biggest one possible since I am counting down from the top
The problem is that you might have found a palindrome for a large i and a small g. It's possible that there's a larger palindrome that's the product of j and k where:
i > j and
g < k
(I hope this makes sense).
Java Implementation:
public class Palindrome {
public static void main(String[] args)
{ int i, j;
int m = 1;
int k =11;
boolean flag = false;
while (true)
{;
if (flag) j = m + 1;
else j = m;
for (i = k; i > 0; i--)
{
j++;
int number, temp, remainder, sum = 0;
number = temp = (1000 - i) * (1000 - j);
while (number > 0)
{
remainder = number % 10;
number /= 10;
sum = sum * 10 + remainder;
}
if (sum == temp)
{
System.out.println("Max value:"+temp);
return;
}
}
if (flag)
m++;
k=k+11;
flag = !flag;
}
}
}
A word on performance. You have the possibility of duplicating many of the products because you are using a pretty simple nested loop approach. For instance, you start with 999*999 and then 999*998, etc. When the inner loop finishes, you will decrement the outer loop and start again with 998*999, which is the same as 999*998.
Really, what you want to do is start the inner loop with the same value as the current outer loop value. This will eliminate your duplicate operations. Something like this...
for (i = 999; i > 99; i--)
{
for (g = i; g > 99; g--)
{
...
However, as Emilio pointed out, your assumption that the first palindrome you find will be the answer is incorrect. You need to compute the biggest numbers first, obviously. So you should try them in this order; 999*999, 999*998, 998*998, 999*997, 998*997, etc...
Haven't tested it but I think you want something like this (pseudo code):
x = 999;
n = 0;
while (++n <= x)
{
j = x;
k = j - n;
while (j >= k)
{
y = j-- * k;
if (check(y))
stop looking
}
}
I found this article which might help you. It has improved brute force approach.
All the above provided answers are excellent, but still I could not restrict myself from writing the code. The code posted by #thyrgle is absolutely perfect. Only a slight correction which he needs to do is just check which product is the maximum.
The code can be as
int i,j,max=0,temp;
for(i=999;i>=100;i--){
for(j=i;j>=100;j--){
temp=i*j;
if(isPalin(temp) && temp>max){
max=temp;
}
}
}
cout<<max<<"\n";
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int a[6];
void convertToString(int xy){
int i,t=100000;
for(i=0;i<6;i++){
a[i]=xy/t;
xy = xy % t;
t=t/10;
}
}
int check(){
int i;
for(i=0;i<3;i++){
if(a[i]!=a[6-i]){
return 0;
}
}
return 1;
}
void main(){
int x,y,xy,status=0;
int i=0,j=0,p=0;
for(x=999;x>99;x--){
for(y=x;y>99;y--){
xy=x*y;
convertToString(xy);
status = check();
if(status==1){
if(xy>p){
p=xy;
i=x;
j=y;
}
}
}
}
printf("\nTwo numbers are %d & %d and their product is %d",i,j,p);
}
x,y=999,999
k=0
pal=[]
while (y>99):
while (x>=100):
m=x*y
n=x*y
while (n!=0):
k=k*10+(n%10)
n=int(n/10)
if(m==k):
if k not in pal:
pal.append(k)
x=x-1
k=0
else:
y,x=y-1,999
pal.sort()
print(pal)
it gives 906609 as the largest palindrome number