Codeforces546C:Soldier and Cards

Description

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

4

2 1 3

2 4 2

Output

6 2

Input

3

1 2

2 1 3

Output

-1

题意：两个士兵进行纸牌游戏，一个有有K1张卡牌，另外一个有K2张；各自抽出各自卡牌的最上面一张，比较数值的大小。如果a士兵大则，将获得b士兵的较小的卡牌。将大的牌放到牌堆最底部，较小的次之。继续游戏。如果a士兵获胜则输出1，如果b士兵获胜则输出2，如果不能分出胜负则输出-1；

代码实现

#include<iostream>
#include<queue>
#include<cstdio>
using namespace std;
queue<int> num1,num2;   //定义两个优先队列
int main()
{
int n;
int n1,n2,num;
int a,b;
int cnount=0;
cin>>n;

cin>>n1;
for(int i=0;i<n1;i++)
{   cin>>num;
num1.push(num);
}
cin>>n2;
for(int i=0;i<n2;i++)
{
cin>>num;
num2.push(num);
}
//分别向num1和num2压入纸牌

while(true)
{

if(num1.empty()||num2.empty()) //若其中的一个战列为空，即输掉游戏。
break;
a=num1.front();
b=num2.front();
//分别去除队列的最上元素，进行比较
num1.pop();
num2.pop();
if(a>b)
{
num1.push(b);
num1.push(a);
}
else
{
num2.push(a);
num2.push(b);
}
//if语判断输赢，并进行push操作
cnount++;//计数器，累加操作的次数
if(cnount>1e6)//无法判断双方的输赢
{
cnount=-1;
break;
}
}

cout<<cnount<<" ";
if(cnount!=-1)
{
if(num1.empty())
cout<<2<<endl;
else
cout<<1<<endl;
}

return 0;
}