[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[30];
int num[30];
bool vis[1005];
int main() {
        sg[0] = 0;
        for (int i = 1; i <= 25; i++) {
                memset(vis, 0, sizeof(vis));
                for (int j = i - 1; j >= 0; j--) {
                        for (int k = j; k >= 0; k--) {
                                vis[sg[j]^sg[k]] = 1;
                        }
                }
                for (int j = 0; j <= 1000; j++) {
                        if (!vis[j]) {
                                sg[i] = j;
                                break;
                        }
                }
        }
        int TNT;
        scanf("%d", &TNT);
        while (TNT--) {
                int ans = 0;
                int n;
                scanf("%d", &n);
                for (int i = 0; i < n; i++) {
                        scanf("%d", &num[i]);
                        if (num[i] & 1) {
                                ans ^= sg[n - i - 1];
                        }
                }


                int a1 = -1, b1, c1;
                int cnt = 0;
                for (int i = 0; i < n - 1; i++) {
                        if (num[i]) {
                                for (int j = i + 1 ; j < n; j++) {
                                        for (int k = j; k < n ; k++) {
                                                if ((ans ^ sg[n - i - 1]^ sg[n - j - 1] ^ sg[n - k - 1]) == 0) {
                                                        cnt++;
                                                        if (a1 == -1) {
                                                                a1 = i, b1 = j, c1 = k;
                                                        }
                                                }
                                        }
                                }
                        }
                }
                if (cnt == 0) {
                        printf("-1 -1 -1\n");
                } else {
                        printf("%d %d %d\n", a1, b1, c1);
                }
                printf("%d\n", cnt);

        }
}