Replying to myself, I should have printed i*i instead of i near the
end of the code:
printf("%i\n",i*i);Dave On Jun 27, 11:47 pm, Dave <[email protected]> wrote: > @Bhavesh: Check the squares of the integers from > ceiling(sqrt(123456789)) to floor(sqrt(987654321)) to see which ones > contain all nine nonzero digits. Since the sum of the nine nonzero > digits is 45, a satisfactory square will be a multiple of 9, and > therefore, we only need consider the squares of integers that are > multiples of 3. Something like this should do the trick: > > int i, j, k; > for( i = 11112 ; i < 31426 ; i += 3 ) > { > j = i * i; > k = 0; > do > { > k |= 1 << (j % 10); > j /= 10; > } while( j > 0 ); > if( k = 0x3FE ) // 11 1111 1110 in binary. > printf("%i\n",i); > } > > Dave > > On Jun 27, 11:01 pm, Bhavesh agrawal <[email protected]> wrote: > > > > > All the nine digits are arranged here so as to form four square numbers: > > > 9 81 324 576 > > > How would you put them together so as to form single smallest possible > > square number and a single largest possible square number.. > > > 139854276 and 923187456 are the answers given everywhere but how to proceed > > this ??- 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.
