There is something wrong with it. I tried your code like this.
int i, n = 8, a[9] = {-1,3,5,1,2,9,10,8,6}; //a[0] is never touched in the
logic
for(i=1; i<=n ;i++ ) //for ease of understanding starting the
array with 1.
{
if(a[ i ] > n )
continue;
else
a[ a [ i ] ]*=n;
}
int x[2], m = 0;
for ( int i = 1; i <=n ; i++)
{
if(a[ i ] % n == 0)
continue;
else
{
x[m++ ] = i;
if(m == 2) break;
}
}
After first for loop, the array a has
-1, 3, 40, 8, 2, 72, 10, 8, 384
And finally, answer array x has 1 and 4, which is not correct.
Could you elaborate your logic in pseudocode?
Thanks,
Channa
On Thu, Jul 30, 2009 at 11:06 PM, Devi G <devs...@gmail.com> wrote:
> I'm sorry...... misunderstood the problem and tot the missin numbers are to
> be less than n.
> Yet it can be modified to suit the pbm.
>
> Suppose the two missin numbers are greater than n, then m==0 when exitin
> the loop.
> So they will be n+1 and n+2 only.
>
> in case, one of the missin numbers is greater than n, then m==1, and can be
> simply found by subtracting the (array_sum+x[0] ) from (sumof 1 to n+2)
> numbers.
>
>
> On Thu, Jul 30, 2009 at 10:55 PM, Devi G <devs...@gmail.com> wrote:
>
>> for(i=1; i<=n ;i++ ) //for ease of understanding starting the
>> array with 1.
>> {
>> if(a[ i ] > n )
>> continue;
>> else
>> a[ a [ i ] ]*=n;
>> }
>>
>> int x[2], m = 0;
>>
>> for ( int i = 1; i <=n ; i++)
>> {
>> if(a[ i ] % n == 0)
>> continue;
>> else
>> {
>> x[m++ ] = i;
>> if(m == 2) break;
>> }
>> }
>>
>> x[] now contain the two missing numbers in ascending order.
>>
>
>
> >
>
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