POJ 1740分析过程
Alice and Bob decide to play a new stone game.At the beginning of the game they pick n(1<=n<=10) piles of stones in a line. Alice and Bob move the stones in turn.
At each step of the game,the player choose a pile,remove at least one stones,then freely move stones from this pile to any other pile that still has stones.
For example:n=4 and the piles have (3,1,4,2) stones.If the player chose the first pile and remove one.Then it can reach the follow states.
2 1 4 2
1 2 4 2（move one stone to Pile 2）
1 1 5 2（move one stone to Pile 3）
1 1 4 3（move one stone to Pile 4）
0 2 5 2（move one stone to Pile 2 and another one to Pile 3）
0 2 4 3（move one stone to Pile 2 and another one to Pile 4）
0 1 5 3（move one stone to Pile 3 and another one to Pile 4）
0 3 4 2（move two stones to Pile 2）
0 1 6 2（move two stones to Pile 3）
0 1 4 4（move two stones to Pile 4）
Alice always moves first. Suppose that both Alice and Bob do their best in the game.
You are to write a program to determine who will finally win the game.
At each step of the game,the player choose a pile,remove at least one stones,then freely move stones from this pile to any other pile that still has stones.
For example:n=4 and the piles have (3,1,4,2) stones.If the player chose the first pile and remove one.Then it can reach the follow states.
2 1 4 2
1 2 4 2（move one stone to Pile 2）
1 1 5 2（move one stone to Pile 3）
1 1 4 3（move one stone to Pile 4）
0 2 5 2（move one stone to Pile 2 and another one to Pile 3）
0 2 4 3（move one stone to Pile 2 and another one to Pile 4）
0 1 5 3（move one stone to Pile 3 and another one to Pile 4）
0 3 4 2（move two stones to Pile 2）
0 1 6 2（move two stones to Pile 3）
0 1 4 4（move two stones to Pile 4）
Alice always moves first. Suppose that both Alice and Bob do their best in the game.
You are to write a program to determine who will finally win the game.
分析：（first表示先手胜，last表示后手胜）
一堆的情况->first
下面分析两堆的情况：
1,1->last
1,2->first(因为此状态可以转化为last的状态)
1,n->first
2,2->last(无法转化为last的状态)
2,n->first(可以转化为2,2的状态)
3,3->last
猜想：（n,m）当且仅当n==m时后手胜。
分析玩两堆之后便可以把两堆的情况推广到n堆的情况：
对于三堆 a1<=a2<=a3的情况：
先手肯定可以制造出 剩下两堆且n==m的情况，让后手陷于必败的局势。
先手可以a3中取走a3+a1-a2个，然后将剩下的a2-a1个移动到a1堆中，制造出a1==a2的局势，根据我们之前的分析，面对此种情况必败。
接下来分析四堆的情况：
a1<=a2<=a3<=a4.
如果a1==a2&&a3==a4，那么先手必败，因为后手总可以保持这样的局势，总令先手面对的是相等局势，最后先手必败。假若先手从a3中取走k个，并将j个移动到a1，那么后手可以从a4中取走k个，将j个移动到a2堆中。
其他情况先手必胜。
结论：面对n堆，a1,a2,,,,,an；如果a1==a2,a3==a4,,,an-1==an，那么后手胜，否则先手胜。
#include"iostream"
#include"algorithm"
using namespace std;
int main()
{
int n;
while(cin>>n,n)
{
int a[11],i,k=1;
for(i=0;i<n;i++)
cin>>a[i];
sort(a,a+i);
if(n%2==0)
{
for(i=0;i<n-1;i+=2)
if(a[i]!=a[i+1]) { k=0;break;}
}
if(k&&n%2==0)
cout<<0<<endl;
else cout<<1<<endl;
}
return 0;
}
