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
4
2 1 3
2 4 2
output
6 2
input
3
1 2
2 1 3
output
-1

纯链表模拟,怎么判断无解当时没想到,所以开了个很大的循环,如果这个循环内还没算出的话就无解。
 #include <iostream>
#include <fstream>
#include <cstdio>
#include <string>
#include <queue>
#include <vector>
#include <map>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cmath>
#include <ctime>
using namespace std; struct Node
{
int num;
Node * next;
}; int main(void)
{
Node * first_1,* first_2,* last_1,* last_2;
Node * cur,* front;
int n,k_1,k_2,num,count = ; scanf("%d",&n); scanf("%d",&k_1);
for(int i = ;i < k_1;i ++)
{
scanf("%d",&num);
cur = new Node;
cur -> num = num;
cur -> next = nullptr;
if(!i)
{
front = first_1 = cur;
last_1 = cur;
continue;
}
front -> next = cur;
front = cur;
last_1 = cur;
} scanf("%d",&k_2);
for(int i = ;i < k_2;i ++)
{
scanf("%d",&num);
cur = new Node;
cur -> num = num;
cur -> next = nullptr;
if(!i)
{
front = first_2 = cur;
last_2 = cur;
continue;
}
front -> next = cur;
front = cur;
last_2 = cur;
} while(first_1 && first_2)
{
Node * temp_1 = first_1;
Node * temp_2 = first_2; if(temp_1 -> num < temp_2 -> num)
{
Node * cur_1 = new Node;
Node * cur_2 = new Node;
cur_1 -> num = temp_1 -> num;
cur_2 -> num = temp_2 -> num; last_2 -> next = cur_1;
cur_1 -> next = cur_2;
cur_2 -> next = nullptr;
last_2 = cur_2;
}
else
{
Node * cur_1 = new Node;
Node * cur_2 = new Node;
cur_1 -> num = temp_1 -> num;
cur_2 -> num = temp_2 -> num; last_1 -> next = cur_2;
cur_2 -> next = cur_1;
cur_1 -> next = nullptr;
last_1 = cur_1;
} first_1 = first_1 -> next;
first_2 = first_2 -> next;
count ++;
if(count > )
{
puts("-1");
return ;
}
}
if(!first_1)
printf("%d 2\n",count);
else
printf("%d 1\n",count); return ;
}

CF Soldier and Cards (模拟)的更多相关文章

  1. Codeforces Round #304 (Div. 2) C. Soldier and Cards —— 模拟题,队列

    题目链接:http://codeforces.com/problemset/problem/546/C 题解: 用两个队列模拟过程就可以了. 特殊的地方是:1.如果等大,那么两张牌都丢弃 : 2.如果 ...

  2. [CF546C] Soldier and Cards - 模拟

    两个人玩牌,首先两个人都拿出自己手牌的最上面的进行拼点,两张拼点牌将都给拼点赢得人,这两张牌放入手牌的顺序是:先放对方的牌再放自己的.若最后有一个人没有手牌了,那么他就输了,求输出拼点的次数和赢得人的 ...

  3. queue+模拟 Codeforces Round #304 (Div. 2) C. Soldier and Cards

    题目传送门 /* 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 queue容器:模拟上述过程,当次数达到最大值时判断为-1 */ #include <cstdio&g ...

  4. cf 546C Soldier and Cards

    题目链接:C. Soldier and Cards Two bored soldiers are playing card war. Their card deck consists of exact ...

  5. 【CodeForces - 546C】Soldier and Cards (vector或队列)

    Soldier and Cards 老样子,直接上国语吧  Descriptions: 两个人打牌,从自己的手牌中抽出最上面的一张比较大小,大的一方可以拿对方的手牌以及自己打掉的手牌重新作为自己的牌, ...

  6. 【codeforces 546C】Soldier and Cards

    time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...

  7. 队列 Soldier and Cards

    Soldier and Cards 题目: Description Two bored soldiers are playing card war. Their card deck consists ...

  8. Codeforces Round #304 (Div. 2) C. Soldier and Cards 水题

    C. Soldier and Cards Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/546 ...

  9. C - Soldier and Cards

    Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Description Two bo ...

随机推荐

  1. android 开源 + 一些素材网站

    ui 设计工具:http://www.sketchcn.com/ 分类汇总: https://github.com/Trinea/android-open-project 直接拿来用!最火的Andro ...

  2. ios transition translate 闪屏问题总结

    webkit在绘制页面时会将结构分为各种层,当层足够大时就会变成很大的平铺层.这样一来webkit在每次页面结构发生变化时不需要都渲染整个页面而是渲染对应层了,这对渲染速度来说相当的重要.webkit ...

  3. CorelDraw x6【Cdr x6】官方简体中文破解版(64位)安装图文教程、破解注册方法

    国内私募机构九鼎控股打造APP,来就送 20元现金领取地址:http://jdb.jiudingcapital.com/phone.html内部邀请码:C8E245J (不写邀请码,没有现金送)国内私 ...

  4. OpenCV 图像处理学习笔记(一)

    解读IplImage结构 typedef struct _IplImage { int nSize;                    /* IplImage大小 */ int ID;       ...

  5. STM32 使用 printf 发送数据配置方法 -- 串口 UART, JTAG SWO, JLINK RTT

    STM32串口通信中使用printf发送数据配置方法(开发环境 Keil RVMDK) http://home.eeworld.com.cn/my/space-uid-338727-blogid-47 ...

  6. IOS 手机助手及越狱助手推荐

    快装系列 1 快装助手(PC版) http://pc.kuaiapp.cn/ 2 快装越狱助手(PC版)http://www.kuaiapp.cn 3 快装商店(手机版)http://yueyu.ku ...

  7. Java浮点值拒绝服务漏洞危害分析

    By 空虚浪子心 http://www.inbreak.net/ JAVA出了漏洞,CVE-2010-4476,会导致拒绝服务攻击.大家能从公告上,看到这样一段代码,挺长的.意思是只有开发人员写出这样 ...

  8. Android Recovery Ui 分析

    Android  recovery和android本质上是两个独立的rootfs, 仅仅是recovery这个rootfs存在的意义就是为android这个rootfs服务,因此被解释为Android ...

  9. 栈上连续定义的int变量,地址相差12个字节

    在VS2010,进行调试的时候,发现连续定义的int变量,地址相差12个字节.这是为什么? 按照我们的理解,int占用4个字节,应该相差4个字节.这是因为VS2010在Debug模式下,int变量占用 ...

  10. jquery 3D 标签云

    http://www.gbin1.com/technology/jquerynews/20111205tagcloudbyjquery/index.html 相关选项 zoom: 90 初始的缩放度  ...