Codeforces Round #304 (Div. 2) C. Soldier and Cards stl应用

C. Soldier and Cards

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard 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.

Sample test(s)

input

42 1 32 4 2

output

6 2

input

31 22 1 3

output

-1

Note

First sample:

Second sample:

题目很简单，主要就是模拟，状态虽然似乎很多，有10！个，但是因为大多数状态是达不到的，所以可以设定一定的步数，如果还是没有结果，就输出-1就可以了，官方给出的是不超过106种。另外也可以用set来保存状态。queue<int>底层是封装了 == 与<k号的。所以可以存在set里做一个标记。相当于hash了吧。

#define INF9000000000#define EPS(double)1e-9#define mod1000000007#define PI3.14159265358979//*******************************************************************************/#endif#define N 20500#define M 100005#define maxn 205#define MOD 1000000000000000007queue<int> q[2];set<pair<queue<int>,queue<int> > > myset;int main(){    int n, m, k, t;while (S(n) != EOF){    myset.clear();while (!q[0].empty())   q[0].pop();while (!q[1].empty())   q[1].pop();for (int i = 0; i < 2; i++){S(k);for (int j = 0; j < k; j++){S(t);q[i].push(t);}}myset.insert(make_pair(q[0],q[1]));int step = 0;while (!q[0].empty() && !q[1].empty()){int s0 = q[0].front();q[0].pop();int s1 = q[1].front();q[1].pop();if (s0 < s1){q[1].push(s0);q[1].push(s1);}else {q[0].push(s1);q[0].push(s0);}step++;//if (step > 10000){//step = -1;//break;//}            if(myset.count(make_pair(q[0],q[1]))){                step = -1;                break;}myset.insert(make_pair(q[0],q[1]));}if (step == -1){cout << step << endl;}else {if (q[0].empty())cout << step << " 2" << endl;elsecout << step << " 1" << endl;}}return 0;}