题目链接：uva 1378 - A Funny Stone Game

题目大意；两个人玩游戏，对于一个序列，轮流操作，每次选中序列中的i，j，k三个位置要求i

解题思路：首先预处理出各个位置上的SG值，然后对于给定序列，枚举位置转移状态后判断是否为必败态即可。

#include 
   
     #include 
    
      #include 
     
       using namespace std; const int maxn = 30; int n, g[maxn], s[maxn]; int SG (int l) { int vis[1000]; memset(vis, 0, sizeof(vis)); for (int i = 0; i < l; i++) { for (int j = 0; j < l; j++) vis[g[i]^g[j]] = 1; } int ret = -1; while (vis[++ret]); return ret; } void init () { g[0] = 0; for (int i = 1; i < maxn; i++) g[i] = SG(i); } int judge () { int ret = 0; for (int i = 0; i < n-1; i++) { if (s[i]&1) ret ^= g[n-1-i]; } return ret; } void put_ans () { for (int i = 0; i < n-1; i++) { if (s[i] == 0) continue; s[i]--; for (int j = i+1; j < n; j++) { s[j]++; for (int k = j; k < n; k++) { s[k]++; if (judge() == 0) { printf(" %d %d %d\n", i, j, k); return; } s[k]--; } s[j]--; } s[i]++; } } int main () { init(); int cas = 1; while (scanf("%d", &n) == 1 && n) { for (int i = 0; i < n; i++) scanf("%d", &s[i]); printf("Game %d:", cas++); if (judge()) put_ans(); else printf(" -1 -1 -1\n"); } return 0; }