Nim Game

https://leetcode.com/discuss/63725/theorem-all-4s-shall-be-false  太机智

Proof:

the base case: when 

n = 4

, as suggested
by the hint from the problem, no matter which number that that first player, the second player would always be able to pick the remaining number.

For 

1* 4 < n < 2 * 4, (n = 5, 6, 7)

,
the first player can reduce the initial number into 4 accordingly, which will leave the death number 4 to the second player. i.e. The numbers 5, 6, 7 are winning numbers for any player who got it first.

Now to the beginning of the next cycle, 

n
= 8

, no matter which number that the first player picks, it would always leave the winning numbers (5, 6, 7) to the second player. Therefore, 8 % 4 == 0, again is a death number.

Following the second case, for numbers between (2*4 = 8) and (3*4=12), which are 

9,
10, 11

, are winning numbers for the first player again, because the first player can always reduce the number into the death number 8.

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

}
};