C. Soldier and Cards

Time Limit: 20 Sec  Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/546/problem/C

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

4
2 1 3
2 4 2

Sample Output

6 2

HINT

题意

玩牌,每次从牌顶上拿出两张牌,然后比大小,然后先扔小的都扔进大的底下,再扔大的在大的那堆底下

然后问你谁赢了,花了几步。

是否循环

题解:

啊,暴力暴力

代码:

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
int Num;
char CH[];
//const int inf=0x7fffffff; //нчоч╢С
const int inf=0x3f3f3f3f;
/* inline void P(int x)
{
Num=0;if(!x){putchar('0');puts("");return;}
while(x>0)CH[++Num]=x%10,x/=10;
while(Num)putchar(CH[Num--]+48);
puts("");
}
*/
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
inline void P(int x)
{
Num=;if(!x){putchar('');puts("");return;}
while(x>)CH[++Num]=x%,x/=;
while(Num)putchar(CH[Num--]+);
puts("");
}
//************************************************************************************** queue<int> q1;
queue<int> q2;
int main()
{
int n=read();
int k1=read();
for(int i=;i<k1;i++)
{
int t=read();
q1.push(t);
}
int k2=read();
for(int i=;i<k2;i++)
{
int t=read();
q2.push(t);
}
int flag=;
while(!q1.empty()&&!q2.empty())
{
flag++;
if(flag>)
break;
int ans1=q1.front(),ans2=q2.front();
q1.pop(),q2.pop();
if(ans1>ans2)
{
q1.push(ans2);
q1.push(ans1);
}
else
{
q2.push(ans1);
q2.push(ans2);
}
}
if(q1.empty()||q2.empty())
{
int ans;
if(q1.empty())
ans=;
else
ans=;
//int ans=flag%2==1?1:2;
printf("%d %d\n",flag,ans);
}
else
cout<<"-1"<<endl;
}

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