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.
思路:很简单(之前大一acm宣讲时大神提过的问题,然而并没有什么卵用),每个人每次只能拿1,2或3块石头,最小是1,最大是3.也就是说下一个人一定可以(假设说两人足够聪明)根据上一个人拿的数量来凑出一个4.
结果很容易推演:甲先手取i块,乙必定根据甲的情况取4-i块,然后经过n个回合(甲取,乙取),若还剩下石头(少于4块),则甲最后取完,甲胜.否则不剩下石头,乙最后取完,乙胜.
归根结底就是看总数是否是4的倍数.
bool canWinNim(int n) {
if (n%4 ==0) return false;
else return true;
}
