if ( n <= 9 )
return (10+n);
ret = 0;
while (n>9)
{
found = false;
for ( i = 9 to 2 )
if (n%i == 0)
{ ret = ret * 10 + i; n /= i; found = true; }
if ( !found )
{
print( "Not possible"); return;
}
}
Now sort the digits of ret and return.
However this will work for small numbers. Numbers with 1000 digits will need
different handling.
On 6/10/07, mukesh tiwari <[EMAIL PROTECTED]> wrote:
>
>
> Hi everbody , i am trying to solve this problem
> http://acm.uva.es/p/v105/10527.html but i am not getting any
> efficient algorithm . One algorithm i applied
> while(n>9)
> for i=2 to 9
> if(n%i=0)//then i is factor of n and may be a digit in the number
> store i for further processing
> n=n/i
>
> now the problem with this algorithm is supose we have input 45
> so answer should be 59 as from defintion
> 59 -> 45
>
> but my algorithm is giving the answer is 335
> as 335 -> 45
> but i have to find out minimum number .Kindly help me . Thnkx in
> advance .
>
>
> >
>
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