Two bored soldiers are playing card war. Their card deck consists of exactly n cards, numbered from 1 to n, all values are different. They divide cards between them in some manner, it‘s possible that they have different number of cards. Then they play a "war"-like card game.
The rules are following. On each turn a fight happens. Each of them picks card from the top of his stack and puts on the table. The one whose card value is bigger wins this fight and takes both cards from the table to the bottom of his stack. More precisely, he first takes his opponent‘s card and puts to the bottom of his stack, and then he puts his card to the bottom of his stack. If after some turn one of the player‘s stack becomes empty, he loses and the other one wins.
You have to calculate how many fights will happen and who will win the game, or state that game won‘t end.
Input
First line contains a single integer n (2 ≤ n ≤ 10), the number of cards.
Second line contains integer k1 (1 ≤ k1 ≤ n - 1), the number of the first soldier‘s cards. Then follow k1 integers that are the values on the first soldier‘s cards, from top to bottom of his stack.
Third line contains integer k2 (k1 + k2 = n), the number of the second soldier‘s cards. Then follow k2 integers that are the values on the second soldier‘s cards, from top to bottom of his stack.
All card values are different.
Output
If somebody wins in this game, print 2 integers where the first one stands for the number of fights before end of game and the second one is 1 or 2 showing which player has won.
If the game won‘t end and will continue forever output - 1.
Sample Input
Input
42 1 32 4 2
Output
6 2
Input
31 22 1 3
Output
-1
解题思路：这个题目我们用队列来解决较好，我们先建立两个队列。解题的关键是我们每次都从两个队列中取出第一个元素，我们判断他们的大小，记住每次都要把队首的元素删除掉；把小的队的元素插入大的队的队尾再把他自己的元素插到后面，知道其中有一个队成为空的。，其中没有空的队就是胜出者。如果经过很多次的循环都没有对变成空的。这是就表示陷入循环，此时输出-1.
程序代码：
#include <iostream>
#include <queue>
using namespace std;
int main()
{
queue<int>v;
queue<int>w;
int n,x,y,c,t,k=0,first=1;
cin>>n;
cin>>x;
while(x--)
{
cin>>c;
v.push(c);
}
cin>>y;
while(y--)
{
cin>>c;
w.push(c);
}
while(!v.empty()&&!w.empty())
{
if(k>1000)
{
first=0;
break;
}
if(v.front()>w.front())
{
v.push(w.front());
v.push(v.front());
v.pop();
w.pop();
k++;
}
else
{
w.push(v.front());
w.push(w.front());
w.pop();
v.pop();
k++;
}
}
t=(v.empty()?2:1);
if(first)
cout<<k<<" "<<t<<endl;
else
cout<<"-1"<<endl;
return 0;
}