先找边界：SG[0] = 0表示当石子数为0时，先手输

然后套SG即可

 

若为n堆，每堆的SG异或即可

class Solution {
public:
    int SG[100001];
    int vis[100];



    bool winnerSquareGame(int n) {


        SG[0] = 0;
        for(int i = 1; i <= n; i++)
        {
            memset(vis, 0, sizeof(vis));
            for(int j = 1; j * j <= i; j++) vis[SG[i - j * j]] = 1;
            for(int j = 0; ; j++)
                if(!vis[j])
                {
                    SG[i] = j;
                    break;
                }
        }
        return SG[n];


    }
};