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/problem/C
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
4
2 1 3
2 4 2
Sample Output
6 2
HINT
题意
玩牌,每次从牌顶上拿出两张牌,然后比大小,然后先扔小的都扔进大的底下,再扔大的在大的那堆底下
然后问你谁赢了,花了几步。
是否循环
题解:
啊,暴力暴力
代码:
//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
int Num;
char CH[];
//const int inf=0x7fffffff; //нчоч╢С
const int inf=0x3f3f3f3f;
/* inline void P(int x)
{
Num=0;if(!x){putchar('0');puts("");return;}
while(x>0)CH[++Num]=x%10,x/=10;
while(Num)putchar(CH[Num--]+48);
puts("");
}
*/
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
inline void P(int x)
{
Num=;if(!x){putchar('');puts("");return;}
while(x>)CH[++Num]=x%,x/=;
while(Num)putchar(CH[Num--]+);
puts("");
}
//************************************************************************************** queue<int> q1;
queue<int> q2;
int main()
{
int n=read();
int k1=read();
for(int i=;i<k1;i++)
{
int t=read();
q1.push(t);
}
int k2=read();
for(int i=;i<k2;i++)
{
int t=read();
q2.push(t);
}
int flag=;
while(!q1.empty()&&!q2.empty())
{
flag++;
if(flag>)
break;
int ans1=q1.front(),ans2=q2.front();
q1.pop(),q2.pop();
if(ans1>ans2)
{
q1.push(ans2);
q1.push(ans1);
}
else
{
q2.push(ans1);
q2.push(ans2);
}
}
if(q1.empty()||q2.empty())
{
int ans;
if(q1.empty())
ans=;
else
ans=;
//int ans=flag%2==1?1:2;
printf("%d %d\n",flag,ans);
}
else
cout<<"-1"<<endl;
}
Codeforces Round #304 (Div. 2) C. Soldier and Cards 水题的更多相关文章
- 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 ...
- Codeforces Round #304 (Div. 2) B. Soldier and Badges 水题
B. Soldier and Badges Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/54 ...
- 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 ...
- DP+埃氏筛法 Codeforces Round #304 (Div. 2) D. Soldier and Number Game
题目传送门 /* 题意:b+1,b+2,...,a 所有数的素数个数和 DP+埃氏筛法:dp[i] 记录i的素数个数和,若i是素数,则为1:否则它可以从一个数乘以素数递推过来 最后改为i之前所有素数个 ...
- 贪心 Codeforces Round #304 (Div. 2) B. Soldier and Badges
题目传送门 /* 题意:问最少增加多少值使变成递增序列 贪心:排序后,每一个值改为前一个值+1,有可能a[i-1] = a[i] + 1,所以要 >= */ #include <cstdi ...
- 水题 Codeforces Round #304 (Div. 2) A. Soldier and Bananas
题目传送门 /* 水题:ans = (1+2+3+...+n) * k - n,开long long */ #include <cstdio> #include <algorithm ...
- 数学+DP Codeforces Round #304 (Div. 2) D. Soldier and Number Game
题目传送门 /* 题意:这题就是求b+1到a的因子个数和. 数学+DP:a[i]保存i的最小因子,dp[i] = dp[i/a[i]] +1;再来一个前缀和 */ /***************** ...
- Codeforces Round #297 (Div. 2)A. Vitaliy and Pie 水题
Codeforces Round #297 (Div. 2)A. Vitaliy and Pie Time Limit: 2 Sec Memory Limit: 256 MBSubmit: xxx ...
- Codeforces Round #290 (Div. 2) A. Fox And Snake 水题
A. Fox And Snake 题目连接: http://codeforces.com/contest/510/problem/A Description Fox Ciel starts to le ...
随机推荐
- C++ 模板特化以及Typelist的相关理解
近日,在学习的过程中第一次接触到了Typelist的相关内容,比如Loki库有一本Modern C++ design的一本书,大概JD搜了一波没有译本,英文版600多R,瞬间从价值上看到了这本书的价值 ...
- http状态码说明
在学习网页设计的时候都应该知道状态码,但我们常见的状态码都是200,404,下面介绍其他的状态值 1开头的http状态码表示临时响应并需要请求者继续执行操作的状态代码. 100 (继续) 请求者应 ...
- linux 实现自动创建ftp用户并创建文件夹
创建一个 createuser.sh的脚本文件 #!/bin/sh #传入的文件名 name=$1 #创建该用户所对应的ftp文件夹 /srv/ftp是我的ftp服务器的根目录 mkdir /sr ...
- nginx 服务器篇
Nginx 服务器类型 1. Web服务器 Web服务器用于提供HTTP(包括HTTPS)的访问,例如Nginx.Apache.IIS等. 2. 应用程序服务器 应用程序服务器能够用于应用程序的运行, ...
- java版云笔记(七)之事务管理
事务管理 事务:程序为了保证业务处理的完整性,执行的一条或多条SQL语句. 事务管理:对事务中的SQL语句进行提交或者回滚. 事物管理对于企业应用来说是至关重要的,好使出现异常情况,它也可以保证数据的 ...
- Nginx1.8.1 编译扩展https
nginx无缝编译扩展https 本贴只限用于通过编译安装的nginx,如果用的是yum源安装请卸载后参见 http://www.cnblogs.com/rslai/p/7851220.html 安装 ...
- Java显式锁学习总结之四:ReentrantLock源码分析
概述 ReentrantLock,即重入锁,是一个和synchronized关键字等价的,支持线程重入的互斥锁.只是在synchronized已有功能基础上添加了一些扩展功能. 除了支持可中断获取锁. ...
- python爬取网易云音乐歌单音乐
在网易云音乐中第一页歌单的url:http://music.163.com/#/discover/playlist/ 依次第二页:http://music.163.com/#/discover/pla ...
- Binary Tree Zigzag Level Order Traversal——关于广度优先的经典面试题
Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to ...
- Hadoop HDFS 单节点部署方案
初学者,再次记录一下. 确保Java 和 Hadoop已安装完毕(每个人的不一定一样,但肯定都有数据,仅供参考) [root@jans hadoop-2.9.0]# pwd /usr/local/ha ...