help in fixing this code..

i need to write a code that flips a number and sums it with the previous form of the number,till i get a number that if i will flip it
i will get the same number
i need to output the number of steps which took the program to transform the input number into the number that if we flip it ,it will remain the same.
for example if i input

679 + 976 = 1655
1655 + 5561 = 7216
7216 + 6127 = 13343
13343 + 34331 = 47674
4 steps

i wrote the code which flips i put a loop over it but its not working
??

Code:

---


int main(){
    int sum=0;
    int counter=0;
        int i;
        int n = 0;
        scanf("%d",&i);
    counter=0;

while(n!=i) {//start while

  while(i)
  {
      n = 10 * n + i % 10;
      i /= 10;
  }
n=i+n;

counter++;
}//end while
printf("%d %d",n,counter);
return 0;
}

---

Hmmm I remember seeing this question some time last week. Perhaps doing a forum search would benefit you.

it was me
i ask about the operation of flipping
i solved it but this is the main problem

Why not just take the number as a string, then flip the string? I think its computationally less complex since theoretically you have it as a string to begin with.

i need to use only integers

Maybe you need to SAVE the original number, as I'm sure after the loop, i is always zero.

--
Mats

Thats a small problem.

my main problem is making this process of summing and flipping like i showed in the example.
i tried to build this proccess in this code but its not working.

Code:

---

int main(){
    int sum=0;
    int counter=0;
        int i;
        int n = 0;
        scanf("%d",&i);
    counter=0;

while(n!=i) {//start while

  while(i)
  {
      n = 10 * n + i % 10;
      i /= 10;
  }
n=i+n;

counter++;
}//end while
printf("%d %d",n,counter);
return 0;
}

---

A small problem that you haven't fixed, as it looks. Do you want me to repeat it: "the variable i is zero after your inner while-loop".

--
Mats

Using too many variables and loops won't get you anywhere. Here's how it will look like in pseudo-code.

Code:

---

n = 0;
read i
while (i is not equal to zero)
{
    n = 10 * n + i % 10
    i /= 10
}
print n

---

Quote:

---

Originally Posted by matsp

A small problem that you haven't fixed, as it looks. Do you want me to repeat it: "the variable i is zero after your inner while-loop".

--
Mats

---

I need to reset the value of variable "i" to its original value(the one i entered the scanf)
??

Theoreticly i want to flip each time the sum of the number and its flipped form
that why i did
i=i+n;

But the original value of "i" needs to be changes each time by sum
so i cant put each time the scanf value in "i".

?

No, you need to SAVE the value before your invert it, and then use the saved value after the loop [which may be roughly the same as what you said, but it's not the value you ENTERED that you need to save, but the previous n from the sum - or the initial value].

You obviously also need to check if the reversed value is equal to the current value, and if so, you are done. Currently, you check it after you have added the values together, which won't work - you'll go one step too far [if you are lucky, it will still get a number that is the same backwards - but not always].

--
Mats

i changed the code like you said
still not working
??

Code:

---

int main(){
    int sum=0;
    int counter=0;
        int i;
        int n = 0;
        scanf("%d",&i);
    counter=0;

do {//start while
  while(i)
  {
      n = 10 * n + i % 10;
      i /= 10;
  }


if (n!=i){
n=i+n;
counter++;
}

}while(n!=i);//end while
printf("%d %d",n,counter);
return 0;
}

---

I tried it with for loop

not working
??

Code:

---

#include <stdio.h>



int main(){

        int num;  //a number
        int f_num = 0;  //flipped number
        scanf("&#37;d",&num);

for(f_num=0;f_num!=num;f_num=f_num+num){

  while(num)
  {
      f_num = 10 *f_num + num % 10;
      num /= 10;
  }
}
printf("%d",f_num);
return 0;
}

---

ok i understand that i need to keep the num value.
I tried to do it here.Its not working.

Code:

---

#include <stdio.h>



int main(){
int count=0;
        int num;  //a number
        int temp;
        int f_num = 0;  //flipped number
        scanf("&#37;d",&num);

do{
if ((num!=f_num)&&(count>0)){
 num=num+f_num;
 temp=num;
 printf("num=%d f_num=%d temp=%d count=%d\n",num,f_num,temp,count);
}
temp=num;
  while(num)
  {
      f_num = 10 *f_num + num % 10;
      num /= 10;
  }
  count++;
  printf("num=%d f_num=%d temp=%d count=%d\n",num,f_num,temp,count);
  num=temp;
  printf("num=%d f_num=%d temp=%d count=%d\n",num,f_num,temp,count);
}while (num!=f_num);
printf("%d",f_num);
return 0;
}

---

I'll take a look at it. Shouldn't be tough, but you never know. :)

I tried to trace it.
and it works on the first loop run.
then on the second loop it give a number which its not supposed.

http://img372.imageshack.us/img372/8783/96751135or0.gif

Code:

---

#include <stdio.h>



int main(){
int count=0;
        int num;  //a number
        int temp;
        int f_num = 0;  //flipped number
        scanf("&#37;d",&num);

do{
if ((num!=f_num)&&(count>0)){
 num=num+f_num;
 temp=num;
 printf("num=%d f_num=%d temp=%d count=%d\n",num,f_num,temp,count);
}
temp=num;
  while(num)
  {
      f_num = 10 *f_num + num % 10;
      num /= 10;
  }
  count++;
  printf("num=%d f_num=%d temp=%d count=%d\n",num,f_num,temp,count);
  num=temp;
  printf("num=%d f_num=%d temp=%d count=%d\n",num,f_num,temp,count);
}while (num!=f_num);
printf("%d",f_num);
return 0;
}

---

I don't even know where to start to try and help you with your code. Try this:

Code:

---

#include <stdio.h>

int flip (int num);

int main (void) {
  int i, j, c = 0;

  fprintf(stdout, "\nEnter the number to flip: ");
  fscanf(stdin, "%d", &i);

  while (i != flip(i)) {
    j = flip(i);
    fprintf(stdout, "%d + %d = %d\n", i, j, i+j);
    c++;
    i += j;
  }

  fprintf(stdout, "Completed in %d steps.\n", c);
  return 0;
}

int flip (int num) {
  int digit, result = 0;

  while (num != 0) {
    digit = num % 10;
    result = result * 10 + digit;
    num /= 10;
  }

  return result;
}

---

Using the original posted code as the template:

Code:

---

/*flip (reverse) the number entered and add it to the last tally. Count the
steps needed (and numbers), until you reach a sum that is the reverse
of itself:

eg: 125 entered: 125+521=646,
646 == it's reverse, so only 1 step was needed.

679 + 976 = 1655
1655 + 5561 = 7216
7216 + 6127 = 13343
13343 + 34331 = 47674
4 steps
*/

#include <stdio.h>

int a[10] = { 0 };

int main()  {
  int count=0, i, j, bad;
  unsigned long int temp, sum;
  unsigned long int num ;     
  unsigned long int f_num = 0;  //flipped number
  printf("\n\n\n Enter a Number: ");
  scanf("&#37;U",&num);
 
  do{
      num=num+f_num;
      f_num = 0;
      temp=num;
      while(temp)  {                   
        f_num = 10 *f_num + temp % 10;
        temp /= 10;
      }
      count++;
      printf("\n num=%lu f_num=%lu count=%d",num,f_num,count);

      //flip the sum into array a[]
      sum = num + f_num;
      temp = sum;
      for(i = 0; i < 10; i++)
        a[i] = 0;
      i = 0;
      while(temp)  {
        a[i] = 10 * a[i] + temp % 10;
        i++;
        temp /= 10;
      }
      i -= 1;
      for(j = 0, bad = 0;j < i; j++, i--)  {
        if(a[j] != a[i]) {
            bad = 1;
            break;
        }
      }
  } while (bad);

  printf("  sum: %lu", sum);

  getch();
  return 0;
}

---

i cant use a function call
so i need to this all in the main function

i tried to change it
but its not working

Code:

---

#include <stdio.h>



int main (void) {
  int i, j, c = 0;
int digit, result = 0;
  printf(  "\nEnter the number to flip: ");
  scanf( "&#37;d", &i);

  while (i != result) {





  while (i != 0) {
    digit = i % 10;
    result = result * 10 + digit;
    i /= 10;
  }

    j=result;
    printf("%d + %d = %d\n", i, j, i+j);
    c++;
    i += j;
  }

  printf( "Completed in %d steps.\n", c);
  return 0;
}

---

Quote:

---

Originally Posted by Adak

Using the original posted code as the template:

Code:

---

/*flip (reverse) the number entered and add it to the last tally. Count the
steps needed (and numbers), until you reach a sum that is the reverse
of itself:

eg: 125 entered: 125+521=646,
646 == it's reverse, so only 1 step was needed.

679 + 976 = 1655
1655 + 5561 = 7216
7216 + 6127 = 13343
13343 + 34331 = 47674
4 steps
*/

#include <stdio.h>

int a[10] = { 0 };

int main()  {
  int count=0, i, j, bad;
  unsigned long int temp, sum;
  unsigned long int num ;     
  unsigned long int f_num = 0;  //flipped number
  printf("\n\n\n Enter a Number: ");
  scanf("%U",&num);
 
  do{
      num=num+f_num;
      f_num = 0;
      temp=num;
      while(temp)  {                   
        f_num = 10 *f_num + temp % 10;
        temp /= 10;
      }
      count++;
      printf("\n num=%lu f_num=%lu count=%d",num,f_num,count);

      //flip the sum into array a[]
      sum = num + f_num;
      temp = sum;
      for(i = 0; i < 10; i++)
        a[i] = 0;
      i = 0;
      while(temp)  {
        a[i] = 10 * a[i] + temp % 10;
        i++;
        temp /= 10;
      }
      i -= 1;
      for(j = 0, bad = 0;j < i; j++, i--)  {
        if(a[j] != a[i]) {
            bad = 1;
            break;
        }
      }
  } while (bad);

  printf("  sum: %lu", sum);

  getch();
  return 0;
}

---

---

i cant use array
and getch()

this is the working code but i need to replace the function calls

Code:

---

#include <stdio.h>



int main (void) {
  int i, j, c = 0;
int digit, result = 0;
  printf(  "\nEnter the number to flip: ");
  scanf( "&#37;d", &i);

  while (i != result) {





  while (i != 0) {
    digit = i % 10;
    result = result * 10 + digit;
    i /= 10;
  }

    j=result;
    printf("%d + %d = %d\n", i, j, i+j);
    c++;
    i += j;
  }

  printf( "Completed in %d steps.\n", c);
  return 0;
}

---

Code:

---

#include <stdio.h>

int flip (int num);

int main (void) {
  int i, j, c,num = 0;
int digit, result = 0;

  printf(  "\nEnter the number to flip: ");
  scanf( "%d", &num);

  while (num != 0) {
    digit = num % 10;
    result = result * 10 + digit;
    num /= 10;
  }


  while (i != result) {

        while (num != 0) {
    digit = num % 10;
    result = result * 10 + digit;
    num /= 10;
  }

    j = result;

    printf("%d + %d = %d\n", i, j, i+j);
    c++;
    i += j;
  }

  printf( "Completed in %d steps.\n", c);
  return 0;
}

---