bzoj 1188

博弈的题目做的还是太少啦。。。 不会写啊啊啊

思路：将每个石子看成一个游戏， 那么整个游戏sg值就是全部石子sg值的异或。

 1 #include<bits/stdc++.h>
 2 #define LL long long
 3 #define fi first
 4 #define se second
 5 #define mk make_pair
 6 #define pii pair<int,int>
 7 #define piii pair<int, pair<int,int>>
 8 
 9 using namespace std;
10 
11 const int N=1e5 + 7;
12 const int M=1e4 + 7;
13 const int inf = 0x3f3f3f3f;
14 const LL INF = 0x3f3f3f3f3f3f3f3f;
15 const int mod = 1e9 + 7;
16 const double eps = 1e-10;
17 const double PI = acos(-1);
18 
19 int n, a[22], sg[22], mark[500];
20 
21 int getSg(int x) {
22     if(sg[x] != -1) return sg[x];
23     memset(mark, 0, sizeof(mark));
24     for(int i = 0; i < x; i++) {
25         for(int j = 0; j <= i; j++) {
26             mark[getSg(i) ^ getSg(j)] = 1;
27         }
28     }
29     for(int i = 0; ; i++)
30         if(!mark[i]) return sg[x] = i;
31 }
32 int main() {
33     memset(sg, -1, sizeof(sg));
34     sg[0] = 0;
35     for(int i = 1; i <= 21; i++) {
36         sg[i] = getSg(i);
37     }
38     int T; scanf("%d", &T);
39     while(T--) {
40         scanf("%d", &n);
41         int ans = 0;
42         for(int i = 0; i < n; i++) {
43             scanf("%d", &a[i]);
44             a[i] %= 2;
45             ans ^= a[i] * sg[n - i - 1];
46         }
47         int tot = 0;
48         for(int i = 0; i < n; i++) {
49             for(int j = i + 1; j < n; j++) {
50                 for(int k = j; k < n; k++) {
51                     if(!(ans ^ sg[n - i - 1] ^ sg[n - j - 1] ^ sg[n - k - 1])) {
52                         if(++tot == 1) printf("%d %d %d\n", i, j, k);
53                     }
54                 }
55             }
56         }
57         if(!tot) puts("-1 -1 -1");
58         printf("%d\n", tot);
59     }
60     return 0;
61 }
62 /*
63 */