ZOJ 2083 Win the Game（SG函数）题解

题意：给一端n块的板，两人玩，每次能涂相邻两块没涂过的板，不能涂的人为输，先手赢输出yes

思路：sg函数打表，练习题

代码：

 #include<queue>
 #include<cstring>
 #include<set>
 #include<map>
 #include<stack>
 #include<cmath>
 #include<vector>
 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #define eps 1e-9
 typedef long long ll;
 const int maxn =  + ;
 const int seed = ;
 const ll MOD = 1e9 + ;
 const int INF = 0x3f3f3f3f;
 using namespace std;
 int sg[], s[];
 int main(){
     sg[] = , sg[] = , sg[] = ;
     for(int i = ; i <= ; i++){
         memset(s, , sizeof(s));
         for(int j = ; j <= i - ; j++){
             int l = j - ;
             int r = i - (j + );
             s[sg[l] ^ sg[r]] = ;
         }
         for(int j = ; j < ; j++){
             if(!s[j]){
                 sg[i] = j;
                 break;
             }
         }
     }
     int n;
     while(~scanf("%d", &n)){
         int ans = ;
         int u;
         while(n--){
             scanf("%d", &u);
             ans ^= sg[u];
         }
         if(ans == ) printf("No\n");
         else printf("Yes\n");
     }
     return ;
 }