【codeforces 546C】Soldier and Cards

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

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.

Examples

input

4

2 1 3

2 4 2

output

6 2

input

3

1 2

2 1 3

output

-1

Note

First sample:

Second sample:

【题目链接】:http://codeforces.com/contest/546/problem/C

【题解】



用vector+reverse来模拟这个过程

然后用map来判断有没有出现循环节;

(用队列,然后轮数到了一个很大的数字还没出结果就直接跳出，这样也可以)

ps:貌似循环1e8次也不会超1s的时限.



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

//const int MAXN = x;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

vector <int> a,b;
int n,r=0;
int k1,k2;
map <pair< vector<int>,vector <int> > ,int> dic;

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    rei(n);
    rei(k1);
    rep1(i,1,k1)
        {
            int x;
            rei(x);
            a.pb(x);
        }
    rei(k2);
    rep1(i,1,k2)
        {
            int x;
            rei(x);
            b.pb(x);
        }
    dic[mp(a,b)] = 1;
    dic[mp(b,a)] = 1;
    while (true)
    {
        r++;
        reverse(a.begin(),a.end());reverse(b.begin(),b.end());
        int A = a.back(),B = b.back();
        a.pop_back();b.pop_back();
        reverse(a.begin(),a.end());reverse(b.begin(),b.end());
        if (A<B)
        {
            b.pb(A);
            b.pb(B);
        }
        else//A>B
        {
            a.pb(B);
            a.pb(A);
        }
        if (dic[mp(a,b)])
        {
            puts("-1");
            return 0;
        }
        if (a.empty())
        {
            printf("%d %d\n",r,2);
            return 0;
        }
        else
            if (b.empty())
            {
                printf("%d %d\n",r,1);
                return 0;
            }
        dic[mp(a,b)] = 1;
        dic[mp(b,a)] = 1;
    }
    return 0;
}