Uva 1378

1378 - A Funny Stone Game

Time limit: 3.000 seconds

The funny stone game is coming. There are n piles of stones, numbered with 0, 1, 2,..., n - 1. Two persons pick stones in turn. In every turn, each person selects three piles of stones numbered i, j, k (i < j, j k and at least one stone left in pile i). Then, the person gets one stone out of pile i, and put one stone into pile j and pile k respectively. (Note: if j = 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 integer n ( 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 pile n - 1respectively.

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

 

Output 

For each case, output a line in the format ``Game t: i j k". 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 of i, jand 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

题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4124
题意：略。
思路：分解的方法，《解析一类组合游戏》中的第一题。几句话看了几个小时，这个伤不起。
　　  后来想到了，相同局面的化简，顿时明白了这种分解的方法。
      子局面，和下一个局面。
      堪称经典题。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;

int SG[25];
bool use[120];
void prepare()
{
    int i,j,s;
    SG[1]=0;
    for(i=2;i<=23;i++)
    {
        memset(use,false,sizeof(use));
        for(j=1;j<i;j++)
            for(s=1;s<i;s++)
            {
                use[ SG[j]^SG[s] ]=true;
            }
        for(j=0;;j++)
            if(use[j]==false)
            {
                SG[i]=j;
                break;
            }
    }
}
int main()
{
    int n,i,j,ans,s,T=0;
    int f[25];
    bool flag;
    prepare();
    while(scanf("%d",&n)>0)
    {
        if(n==0)break;
        for(i=0 ,ans=0;i<n;i++)
        {
            scanf("%d",&f[i]);
            if(f[i]&1)
            {
                ans=ans^SG[n-i];
            }
        }
        printf("Game %d: ",++T);
        if(ans==0)
        {
            printf("-1 -1 -1
");
            continue;
        }
        flag=false;
        for(i=0;i<n;i++)
        {
            if(flag==true) break;
            if(f[i]==0)continue;
            for(j=i+1;j<n;j++)
            {
                if(flag==true) break;
                if(f[j]==0)continue;
                for(s=i+1;s<n;s++)
                {
                    if(f[s]==0)continue;
                    if( (SG[n-i]^SG[n-j]^SG[n-s])== ans)
                    {
                        printf("%d %d %d
",i,j,s);
                        flag=true;
                        break;
                    }
                }
            }
        }

    }
    return 0;
}