题目传送门
1 /*
2 题意：两堆牌，每次拿出上面的牌做比较，大的一方收走两张牌，直到一方没有牌
3 queue容器：模拟上述过程，当次数达到最大值时判断为-1
4 */
5 #include <cstdio>
6 #include <iostream>
7 #include <algorithm>
8 #include <cstring>
9 #include <string>
10 #include <stack>
11 #include <cmath>
12 #include <queue>
13 using namespace std;
14
15 const int MAXN = 20;
16 const int INF = 0x3f3f3f3f;
17 int x[MAXN];
18 queue<int> Q1;
19 queue<int> Q2;
20
21 int main(void) //Codeforces Round #304 (Div. 2) C. Soldier and Cards
22 {
23 int n, m1, m2;
24 while (scanf ("%d", &n) == 1)
25 {
26 while (!Q1.empty ()) Q1.pop ();
27 while (!Q2.empty ()) Q2.pop ();
28
29 int x;
30 scanf ("%d", &m1);
31 for (int i=1; i<=m1; ++i)
32 {
33 scanf ("%d", &x); Q1.push (x);
34 }
35
36 scanf ("%d", &m2);
37 for (int i=1; i<=m2; ++i)
38 {
39 scanf ("%d", &x); Q2.push (x);
40 }
41
42 int cnt = 0;
43 while (!Q1.empty () && !Q2.empty ())
44 {
45 int x = Q1.front (); int y = Q2.front ();
46 Q1.pop (); Q2.pop ();
47 if (x < y) {Q2.push (x); Q2.push (y);}
48 else {Q1.push (y); Q1.push (x);}
49 ++cnt;
50 if (cnt == 10000) break;
51 }
52
53 if (cnt == 10000) puts ("-1");
54 else
55 {
56 printf ("%d %d
", cnt, (Q1.empty ()) ? 2 : 1);
57 }
58 }
59
60 return 0;
61 }
62
63 /*
64 4
65 2 1 3
66 2 4 2
67 3
68 1 2
69 2 1 3
70 */