|
|
back to boardvariation --> possible a new problem I was trying to solve this and I kinda mis-understood in the first place, which makes the problem hard to solve. If the player can decide on the first move (just the first move), then it would be much more complex. For example, 5 1 2 3 4 5 3 3 The current answer is 12 3 because the first player takes 3, 4, 5 and the second player takes 0 2 1 However, if we allow players to skip their first pick, in other words, (move and pick) instead of (pick and move). So there are few more possibilities. if first player picks 3, then the second player can pick 4 and 5 ==> 6, 9 if the first player picks 4, then the second player can pick 3, 2 and 1 ==> 9, 6 maybe I think too much ? :) Re: variation --> possible a new problem You helped me to solve this problem. But the result in this test case is 12, 3. :D Edited by author 20.06.2014 10:31 |
|
|