A number which when multiplied by 2 gives its permutation,my code isn't giving any such .

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C++

#include<stdio.h>
#include<math.h>
int numdig(int x)
{
    int i=0;
    for(i=0;i<100;i++)
    {
        x=x/10;
        if(x==0)
            break;
    }
return (i+1);
}
void sort(int a[],int n)
{
    int i=0,j,temp;
    for(i=0;i<n;i++)
    {
        for(j=i+1;j<n;j++)
        {
            if(a[i]>a[j])
            {
               temp=a[i];
                a[i]=a[j];
                a[j]=temp;
            }
        }
    }
}
int numcheck(int x)
{   int count=0;
    int i=0,j,y;
    int a[50],b[50];
    for(i=0;i<50;i++)
    {
        a[i]=x%10;
        x=x/10;
        if(x==0) break;
    }
    j=i; y=2*x;
     for(i=0;i<50;i++)
    {
        b[i]=y%10;
        y=y/10;
    }
    sort(a,j+1);sort(b,j+1);
    for(i=0;i<j;i++)
    {
        if(a[i]==b[i]) count++;
    }
    if(count==j+1) return 1;
    else return 0;
}
int main()
{
  long long int a,b,i; int t,k;
    printf("enter the number of cases\n");
    scanf("%d",&t);

    for(k=0;k<t;k++)
    {
printf("enter the number which are upper bound and lower bound\n");

    scanf("%d",&a);
    scanf("%d",&b);
    for(i=a;i<b;i++)
    {
        if(numdig(i)!=numdig(2i))
            break;
        else if (numcheck(i)==1)
        printf("this is one of the required number : %d",i);
        else continue;
    }
    }
    return 0;

}

Posted 9-May-15 2:19am

Member 11324568

Updated 9-May-15 4:05am

v3

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Comments

Richard MacCutchan 9-May-15 8:47am

You need to provide more details. What is this supposed to do, what does it actually do and why is that wrong? And please indicate where the error occurs.

Member 11324568 9-May-15 8:50am

first input which i defined as number of cases is actually the number of different ranges between which you want to check.And the other two inputs are the ranges for each of the case.I need to find a number x such that x and 2x are permutations of eachother.I need to print all such x between the different sets of ranges in each of the cases.Thanks...there is no error but any nummber isn't being printed..so i am confused if my code is logically incorrect or really there is no such number between the ranges i input...thanks

Richard MacCutchan 9-May-15 8:54am

I am afraid that does not help. You need to use your debugger, or add some extra print commands, to see what is happening at different stages of the program.

CPallini 9-May-15 9:20am

What does it mean: "such that x and 2x are palindromes of each other"? Can you please give us an example?

Member 11324568 9-May-15 10:07am

i'm sorry @cpallini its permutation not palindrome

Frankie-C 10-May-15 6:54am

See http://en.wikipedia.org/wiki/142857_%28number%29, Repeating decimal and Cyclic numbers.

2 solutions

Solution 1

Start by thinking about what a Palindrome is: it's a word or phrase that when reversed is the same as it was to start with: "TACO CAT" for example.
So what you are looking for is a number like "12345" that when doubled equals it's own reverse: "54321" (clearly, that one doesn't work).

So start by writing a function to return the palindrome of a number: pass it an integer (12345, say) and it returns it's reverse (54321 this time). You can then use that to compare and see if it's the same number.

Since this is your homework, I won't give you the code. But I'll give you a clue or two:
1) The modulus operator "%" will extract the least significant decimal digit if you use 10.
2) The integer divide operator "/" will "throw away" the least significant digit if you use 10.
3) Looping until a value is zero is really, really, easy...

Give it a try: if you meet a problem and you can't solve it, feel free to ask about it.

Posted 9-May-15 3:51am

OriginalGriff

Comments

Member 11324568 9-May-15 10:06am

i'm extremely sorry sir.It is actually permutation i got confused with the two words..Its not a homework sir,i took this problem from codechef.com/hardquestions section

Member 11324568 9-May-15 10:09am

what i did is i extracted all the digits of the number x and 2x into two different arrays and then sorted those two arrays and checked for the equality of the two arrays...for it i initially created function which returns the number of digits ,if they are equal then i checked furthur by checking the digits else i broke the loop..

OriginalGriff 9-May-15 10:14am

I think you need to read the question really carefully - permutation is something else again:
"In mathematics, the notion of permutation relates to the act of rearranging, or permuting, all the members of a set into some sequence or order (unlike combinations, which are selections of some members of the set where order is disregarded)." -- Thanks Google!

If what you mean is that you are looking for a number that when doubled has exactly the same digits (but in a different order) then I'd cheat: Sort the digits of each number and compare the sorted values...

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CPallini · Accepted Answer · 2015-05-09T05:08:00

Your ideas are good, however the implementation is not so good.
The following code ( please note: I don't claim it is correct :-) ) produces

found 125874 (251748)

as first of such numbers.

C

#include<stdio.h>

void sort(int a[],int n)
{
  int i, j, temp;

  for(i=0;i<n-1;i++)
  {
    for(j=i+1;j<n;j++)
    {
      if ( a[i] > a[j] )
      {
        temp = a[i];
        a[i] = a[j];
        a[j]=temp;
      }
    }
  }
}

int make_array(int x, int a[], int size)
{
  int i;

  for (i = 0; i < size; ++i)
  {
    a[i] = x % 10;
    x /= 10;
    if ( x == 0 ) break;
  }
  return (i +1);
}

int numcheck(int x)
{
  int i;
  int sa, sa2;
  int a[50], a2[50];

  sa  =  make_array( x, a, 50);
  sa2 =  make_array( x * 2, a2, 50);
  if ( sa != sa2) return 0;

  sort( a, sa);
  sort( a2, sa);

  for(i=0; i<sa; i++)
  {
    if ( a[i] != a2[i] ) break;
  }
  if (i != sa) return 0;
  return 1;
}

int main()
{
  int a,b,i; int t,k;
  printf("enter the number of cases\n");
  scanf("%d",&t);

  for(k=0;k<t;k++)
  {
    printf("enter the number which are upper bound and lower bound\n");

    scanf("%d",&a);
    scanf("%d",&b);
    for(i=a;i<b;i++)
    {
      if ( numcheck( i ) == 1)  printf("found %d (%d) \n", i, 2*i);
    }
  }
  return 0;
}