blaat submitted Solution 227415 to Problem 627. Compute a dot product of two vectors x and y
on 5 Apr 2013
blaat submitted Solution 227411 to Problem 528. Find the largest value in the 3D matrix
on 5 Apr 2013
blaat submitted Solution 227400 to Problem 410. Back to basics 20 - singleton dimensions
on 5 Apr 2013
blaat submitted Solution 225740 to Problem 411. Back to basics 21 - Matrix replicating
on 30 Mar 2013
blaat submitted Solution 225737 to Problem 731. Given a window, how many subsets of a vector sum positive
on 30 Mar 2013
blaat submitted Solution 225732 to Problem 413. Back to basics 23 - Triangular matrix
on 30 Mar 2013
blaat submitted Solution 225729 to Problem 353. Back to basics 9 - Indexed References
on 30 Mar 2013
blaat submitted Solution 225726 to Problem 649. Return the first and last character of a string
on 30 Mar 2013
blaat submitted Solution 147854 to Problem 492. Find best placement for ordered dominoes (harder)
on 13 Oct 2012
blaat submitted Solution 147849 to Problem 487. Find perfect placement of non-rotating dominoes (easier)
on 13 Oct 2012
blaat submitted Solution 42825 to Problem 246. Project Euler: Problem 8, Find largest product in a large string of numbers
on 16 Feb 2012
blaat submitted a Comment to Solution 41338
What it does: it finds the maximum number of occurrences for each prime number smaller than x in the factorisations of the numbers 1:x (e.g., for x = 10, the maximum number of occurrences for 2 is 3, since 2*2*2 = 8). If the product of all prime factors taken to the power of their maximum # of occurrences is then taken, the smallest number that is divisible by 1:10 is obtained. So for x = 10: 2 * 2 * 2 * 3 * 3 * 5 * 7 = 2520.
on 14 Feb 2012
blaat submitted Solution 41340 to Problem 290. Make one big string out of two smaller strings
on 14 Feb 2012
blaat submitted Solution 41338 to Problem 239. Project Euler: Problem 5, Smallest multiple
on 14 Feb 2012
blaat submitted Solution 41262 to Problem 250. Project Euler: Problem 10, Sum of Primes
on 14 Feb 2012