[leetcode] NimGame

4000

[leetcode] NimGame

You are playing the following Nim Game with your friend: There is a heap of stones on the table, each time one of you take turns to remove 1 to 3 stones. The one who removes the last stone will be the winner. You will take the first turn to remove the stones.

Both of you are very clever and have optimal strategies for the game. Write a function to determine whether you can win the game given the number of stones in the heap.

For example, if there are 4 stones in the heap, then you will never win the game: no matter 1, 2, or 3 stones you remove, the last stone will always be removed by your friend.

-Hint:

If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner?

[原题]

  一开始的思路是：当轮到自己拿石子的时候，石子的数目是4个，就一定输了。这样也许能从底层构造出为让自己必然遇到4个石子的情况，其全为false。如果第n-1把（对方）为4+3，4+2，4+1的情况，则第n把（本方）必为4。由此得出，第n-2把本方应该为4+1+3~4+3+1间的数值。以此类推。

  可以用类似动态规划的备忘录方法计算出第一把的数目范围，但是这个范围到后面会越来越大，不可能表示出来。然后注意到返回true或false的数目的也许会有规律（好吧我开始google了╮(╯▽╰)╭）。

博弈论

博弈论的三个算法

本题属于BashGame

面对[1~3]时必胜；

面对[1~3]+1时必败；

则对于1，…n必胜，n+1为必败，则n+1+1,…n+1+n为必胜；2n+2为必败。

类推得k(n+1)必败。

class Solution {
public:
bool canWinNim(int n) {
if(n%4!=0)
return true;
return false;
}
};