[HNOI2007]分裂游戏 SG打表博弈

结论:其实每一个巧克力都是一堆石子 它的石子数就是它到队尾的距离

打一个SG表即可

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int sg[];
int num[];
bool vis[];
int main() {
        sg[] = ;
        for (int i = ; i <= ; i++) {
                memset(vis, , sizeof(vis));
                for (int j = i - ; j >= ; j--) {
                        for (int k = j; k >= ; k--) {
                                vis[sg[j]^sg[k]] = ;
                        }
                }
                for (int j = ; j <= ; j++) {
                        if (!vis[j]) {
                                sg[i] = j;
                                break;
                        }
                }
        }
        int TNT;
        scanf("%d", &TNT);
        while (TNT--) {
                int ans = ;
                int n;
                scanf("%d", &n);
                for (int i = ; i < n; i++) {
                        scanf("%d", &num[i]);
                        if (num[i] & ) {
                                ans ^= sg[n - i - ];
                        }
                }

                int a1 = -, b1, c1;
                int cnt = ;
                for (int i = ; i < n - ; i++) {
                        if (num[i]) {
                                for (int j = i +  ; j < n; j++) {
                                        for (int k = j; k < n ; k++) {
                                                if ((ans ^ sg[n - i - ]^ sg[n - j - ] ^ sg[n - k - ]) == ) {
                                                        cnt++;
                                                        if (a1 == -) {
                                                                a1 = i, b1 = j, c1 = k;
                                                        }
                                                }
                                        }
                                }
                        }
                }
                if (cnt == ) {
                        printf("-1 -1 -1\n");
                } else {
                        printf("%d %d %d\n", a1, b1, c1);
                }
                printf("%d\n", cnt);

        }
}