2015 HUAS Summer Training#2~B

最新推荐文章于 2015-07-24 21:29:00 发布

转载最新推荐文章于 2015-07-24 21:29:00 发布 · 80 阅读

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原文链接：http://www.cnblogs.com/chenchunhui/p/4667398.html

两位士兵使用编号从1到n的不同卡牌进行对决，遵循特定规则，计算对决次数及胜者。

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

Input

4
2 1 3
2 4 2

Output

6 2

Input

3
1 2
2 1 3

Output

-1
解题思路：本题其实是一个队列问题，输入数据后将它们各自放入自己的队列中，然后一一拿出来进行比较，哪一方的数字比较大就把数字放进哪一个队列中，需要注意的是要先将对方的数放进去再放自己的。当不能判断胜负时又是另一种输出。还要注意的是判断循环继续的条件时要将条件放大点。
程序代码：

#include<cstdio>
#include<queue>
using namespace std;

int main()
{
int n,x,i,t;
queue<int>A,B;
scanf("%d",&n);
if(n>=2&&n<=10)
{
  scanf("%d",&x);
  for(i=1;i<=x;i++)
  {
   int k;
   scanf("%d",&k);
   A.push(k);
  }
     scanf("%d",&t);
  for(i=1;i<=t;i++)
  {
   int k;
   scanf("%d",&k);
   B.push(k);
  }
  int count=0;
  while(!A.empty()&&!B.empty())
  {
   if(count>1000) break;
   else
   {
    int a=A.front();
    int b=B.front();
    if(a>b)
    {
     A.pop();
     B.pop();
     A.push(b);
     A.push(a);
     count++;
    }
    else
    {
     A.pop();
     B.pop();
     B.push(a);
     B.push(b);
     count++;
    }
   }
  }
     if(A.empty())   printf("%d 2\n",count);
  else if(B.empty()) printf("%d 1\n",count);
  else printf("-1\n");
}
return 0;
}

转载于:https://www.cnblogs.com/chenchunhui/p/4667398.html