UVA - 1378 A Funny Stone Game （SG定理）

Description

The funny stone game is coming. There are n piles of stones, numbered with0, 1, 2,...,
n - 1. Two persons pick stones in turn. In every turn, each person selects three piles of stones numberedi,
j, k (i <
j, jk and at least one stone left in pilei).
Then, the person gets one stone out of pilei, and put one stone into pile
j and pile k respectively. (Note: ifj =
k, it will be the same as putting two stones into pilej). One will fail if he can‘t pick stones according to the rule.

David is the player who first picks stones and he hopes to win the game. Can you write a program to help him?

The number of piles, n, does not exceed
23. The number of stones in each pile does not exceed 1000. Suppose the opponent player is very smart and he will follow the optimized strategy to pick stones.

Input

Input contains several cases. Each case has two lines. The first line contains a positive integern (
1 n
23) indicating the number of piles of stones. The second line contains
n non-negative integers separated by blanks, S₀,...S_n-1 (0
S_i
1000), indicating the number of stones in pile 0 to pilen - 1 respectively.

The last case is followed by a line containing a zero.

Output

For each case, output a line in the format `` Game t:ijk".t is the case number.
i,j and
k indicates which three piles David shall select at the first step if he wants to win. If there are multiple groups ofi,
j and k, output the group with the minimized lexicographic order. If there are no strategies to win the game,i,
j and k are equal to
-1.

Sample Input

4
1 0 1 100
3
1 0 5
2
2 1
0

Sample Output

Game 1: 0 2 3
Game 2: 0 1 1
Game 3: -1 -1 -1
题意：有n堆石头，编号为0-n-1，第i堆初始有si个石头，两个游戏者轮流操作，每次可以选3堆i,j,k，使得i<j<=k，且第i堆至少还剩一个石子，然后从第i堆拿走一个石子
再往第j堆和第k堆各放一个石子（j和k可以一样），判断先手必胜还是必败，如果必胜，输出字典序最小的一个操作（i,j,k）。
思路：可以把每个石子看作是独立的游戏，然后我们从左到右重现排列成n-1，n-2.....0,这样方便我们SG状态的转移，所以我们就可以枚举每个i的后继状态sg[j]^sg[k]了，最后判断一下
操作一次后的SG值是不是为0，也就是必败
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 105;

int n, a[maxn], sg[maxn], vis[maxn];

void init() {
	for (int i = 0; i <= 23; i++) {
		memset(vis, 0, sizeof(vis));
		for (int j = 0; j < i; j++)
			for (int k = j; k < i; k++)
				vis[sg[k]^sg[j]] = 1;
		for (int j = 0; ; j++)
			if (!vis[j]) {
				sg[i] = j;
				break;
			}
	}
}

int main() {
	init();
	int cas = 1;
	while (scanf("%d", &n) != EOF && n) {
		int ans = 0;
		for (int i = 0; i < n; i++) {
			scanf("%d", &a[i]);
			if (a[i] & 1)
				ans ^= sg[n-1-i];
		}
		printf("Game %d: ", cas++);
		if (ans == 0)
			printf("-1 -1 -1\n");
		else {
			int flag = 0;
			for (int i = 0; i < n; i++) {
				if (a[i] == 0)
					continue;
				for (int j = i+1; j < n; j++) {
					for (int k = j; k < n; k++)
						if ((ans^sg[n-1-i]^sg[n-1-j]^sg[n-1-k]) == 0) {
							printf("%d %d %d\n", i, j, k);
							flag = 1;
							break;
						}
					if (flag)
						break;
				}
				if (flag)
					break;
			}
		}
	}
	return 0;
}