Codeforces - 546C. Soldier and Cards- Java

http://blog.csdn.net/kenden23/article/details/49027921

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

4
2 1 3
2 4 2

output

6 2

input

3
1 2
2 1 3

output

-1

Note

First sample:

Second sample:

题目不算太难，主要考小算法和代码功力，我是使用Java写的，三个小算法：

1. 使用所谓的two-point方法，用一个数组模拟循环数列，two-point方法简单说就是用前后两点对数组进行定点

2. 模拟游戏去玩，因为数据不大，模拟速度也是很快的

3. 利用hash表的方法记录游戏过的状态，比如记录10个数字组成的数列作为一个状态，可以使用原始的数组作为一个状态，也可以把数字转换成字符串，最好的方法还是直接转换成一个整数记录状态，因为10个数字组合起来成最大的整数不会超过一个长整形

 public static void main(String[] args) {
  Scanner in = new Scanner(System.in);
  int n = in.nextInt(); int k_1; int k_2;
  
  long begin_1 = 0;  
  k_1 = in.nextInt();
  int k_1_arr[] = new int[n];  
  for (int i = 0 ; i < k_1; i++){
   k_1_arr[i] = in.nextInt();
   begin_1 = begin_1 * 10 + k_1_arr[i];
  }
  
  long begin_2 = 0;
  k_2 = in.nextInt();
  int k_2_arr[] = new int[n];
  for (int i = 0; i < k_2; i++){
   k_2_arr[i] = in.nextInt();
   begin_2 = begin_2 * 10 + k_2_arr[i];
  }
    
  int c = 0, front_1 = 0, end_1 = k_1, front_2 = 0, end_2 = k_2;
  int win_side = -1;
  
  HashMap<Long, Long> states = new HashMap<Long, Long>();
  states.put(begin_1, begin_2);
  
  while (true){
   /// judge winner
   if (0 == k_1) {
    win_side = 2;
    break;
   } else if (0 == k_2) {
    win_side = 1;
    break;
   }
   
   /// mimic the game by code!
   ++c;
   if (front_1 >= n) front_1 -= n;
   if (front_2 >= n) front_2 -= n;
   if (k_1_arr[front_1] > k_2_arr[front_2]){
    if (end_1 >= n) end_1 -= n;
    k_1_arr[end_1++] = k_2_arr[front_2++];
    if (end_1 >= n) end_1 -= n;
    k_1_arr[end_1++] = k_1_arr[front_1++];
    k_1++; k_2--;
   } else {
    if (end_2 >= n) end_2 -= n;
    k_2_arr[end_2++] = k_1_arr[front_1++];
    if (end_2 >= n) end_2 -= n;
    k_2_arr[end_2++] = k_2_arr[front_2++];
    k_1--; k_2++;
   }
   
   /// handle not solution, a tie
   long last_1 = 0;
   for (int i = 0, j = front_1; i < k_1; i++, j++){
    if (j >= n) j -= n;
    last_1 = last_1 * 10 + k_1_arr[j];
   }
   long last_2 = 0L;
   for (int i = 0, j = front_2; i < k_2; i++, j++) {
    if (j >= n) j -= n;
    last_2 = last_2 * 10 + k_2_arr[j];
   }
   Long v = states.get(last_1);
   
   if (null != v && v.longValue() == last_2) break;
   states.put(last_1, last_2);
  }
  
  if (-1 == win_side) { System.out.println(-1); }
  else{
   System.out.print(c+' '+win_side);
  }
 }

http://www.lai18.com/content/604806.html

1.     while ( ~scanf ( "%d" , &n ) )  
2.     {  
3.         scanf ( "%d" , &k1 );  
4.         while ( k1-- )  
5.         {  
6.             scanf ( "%d" , &a );  
7.             q1.push ( a );  
8.         }  
9.         scanf ( "%d" , &k2 );  
10.         while ( k2-- )  
11.         {  
12.             scanf ( "%d" , &a );  
13.             q2.push ( a );  
14.         }  
15.         int ans = 0;  
16.         while ( true )  
17.         {  
18.             if ( q1.empty() || q2.empty() ) break;  
19.             int u = q1.front() , v = q2.front();  
20.             q1.pop();  
21.             q2.pop();  
22.             if ( u > v )  
23.             {  
24.                 q1.push ( v );  
25.                 q1.push ( u );  
26.             }  
27.             else   
28.             {  
29.                 q2.push ( u );  
30.                 q2.push ( v );  
31.             }  
32.             ans++;  
33.             if ( ans > 1e6 )   
34.             {  
35.                 ans = -1;  
36.                 break;  
37.             }  
38.         }  
39.         printf ( "%d" , ans );  
40.         if ( ans != -1 )  
41.             printf ( " %d\n" , q1.empty()?2:1);  
42.         else puts("");  
43.     }