@dave But you are going over elements from 35000 to 98000 something anyway How is it O(sqrt n) ? But I don't think any other approach exists .(One is to use permutations (10!-9!) )
On Feb 27, 3:03 am, Dave <[email protected]> wrote: > Another optimization: Since the sum of the digits of n^2 is 45, n^2 is > divisible by 9. Thus, we need consider only n values that are > divisible by 3. Thus, the outer for-loop can be written > > for( n = 31992 ; n < 99381 ; n+=3 ) > > Dave > > On Feb 23, 7:02 pm, Dave <[email protected]> wrote: > > > Try this: > > > int i,k,n; > > long long j,nsq; > > for( n = 31623 ; n < 100000 ; ++n ) > > { > > nsq = (long long)n * (long long)n; > > j = nsq; > > k = 0; > > for( i = 0 ; i < 10; ++i ) > > { > > k |= (1 << (j % 10)); > > j /= 10; > > } > > if( k == 01777 ) > > printf("%i %lli\n",n,nsq); > > } > > > It finds 76 answers in the blink of an eye, the first being 32043^2 > > and the last being 99066^2. > > > Dave > > > On Feb 22, 3:17 pm, bittu <[email protected]> wrote: > > > > How to find a number of 10 digits (non repeated digits) which is a > > > perfect square? perfect square examples: 9 (3x3) 16 (4x4) 25(5x) etc. > > > Ten digit number example 1,234,567,890 > > > > Thanks & Regards > > > Shashank- Hide quoted text - > > > - Show quoted text - -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
