Is Derek lying?

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 441    Accepted Submission(s): 266

Problem Description
Derek and Alfia are good friends.Derek is Chinese,and Alfia
is Austrian.This summer holiday,they both participate in the summer
camp of Borussia Dortmund.During the summer camp,there will be fan tests
at intervals.The test consists of N choice questions and each question is followed by three choices marked “A” “B” and “C”.Each question has only one correct answer and each question is worth 1 point.It means that if your answer for this question is right,you can get 1 point.The total score of a person is the sum of marks for all questions.When the test is over,the computer will tell Derek the total score of him and Alfia.Then Alfia will ask Derek the total score of her and he will tell her: “My total score is X,your total score is Y.”But Derek is naughty,sometimes he may lie to her. Here give you the answer that Derek and Alfia made,you should judge whether Derek is lying.If there exists a set of standard answer satisfy the total score that Derek said,you can consider he is not lying,otherwise he is lying.
 
Input
The first line consists of an integer T,represents the number of test cases.

For each test case,there will be three lines.

The first line consists of three integers N,X,Y,the meaning is mentioned above.

The second line consists of N characters,each character is “A” “B” or “C”,which represents the answer of Derek for each question.

The third line consists of N characters,the same form as the second line,which represents the answer of Alfia for each question.

Data Range:1≤N≤80000,0≤X,Y≤N,∑Ti=1N≤300000

 
Output
For each test case,the output will be only a line.

Please print “Lying” if you can make sure that Derek is lying,otherwise please print “Not lying”.

 
Sample Input
2
3 1 3
AAA
ABC
5 5 0
ABCBC
ACBCB
 
Sample Output
Not lying
Lying
统计答案不一样的数量,所以a,b得分之差小于等于dif,相等是相同的全对,同理x+y<=n*2-dif,相等情况是打出来的全对
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
char a[],b[];
int ans,dif,t,n,la,lb;
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&n,&la,&lb);
scanf("%s%s",&a,&b);
ans=dif=;
for(int i=;i<n;i++)
{
if(a[i]==b[i]) ans++;
else dif++;
}
if(abs(la-lb)>dif || la+lb>n*-dif)puts("Lying");
else puts("Not lying");
}
return ;
}

HDU 多校联合 6045的更多相关文章

  1. HDU 多校联合 6033 6043

    http://acm.hdu.edu.cn/showproblem.php?pid=6033 Add More Zero Time Limit: 2000/1000 MS (Java/Others)  ...

  2. HDU 多校联合练习赛2 Warm up 2 二分图匹配

    Warm up 2 Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Total ...

  3. hdu 5288||2015多校联合第一场1001题

    pid=5288">http://acm.hdu.edu.cn/showproblem.php?pid=5288 Problem Description OO has got a ar ...

  4. HDU 5792---2016暑假多校联合---World is Exploding

    2016暑假多校联合---World is Exploding Problem Description Given a sequence A with length n,count how many ...

  5. 2016暑假多校联合---Windows 10

    2016暑假多校联合---Windows 10(HDU:5802) Problem Description Long long ago, there was an old monk living on ...

  6. hdu5379||2015多校联合第7场1011 树形统计

    pid=5379">http://acm.hdu.edu.cn/showproblem.php? pid=5379 Problem Description Little sun is ...

  7. 2016暑假多校联合---Rikka with Sequence (线段树)

    2016暑假多校联合---Rikka with Sequence (线段树) Problem Description As we know, Rikka is poor at math. Yuta i ...

  8. 2016暑假多校联合---Substring(后缀数组)

    2016暑假多校联合---Substring Problem Description ?? is practicing his program skill, and now he is given a ...

  9. 2016暑假多校联合---To My Girlfriend

    2016暑假多校联合---To My Girlfriend Problem Description Dear Guo I never forget the moment I met with you. ...

随机推荐

  1. Java 中的事件监听机制

    看项目代码时遇到了好多事件监听机制相关的代码.现学习一下: java事件机制包含三个部分:事件.事件监听器.事件源. 1.事件:继承自java.util.EventObject类,开发人员自己定义. ...

  2. Android-Volley网络通信框架(StringRequest &amp; JsonObjectRequest)

    1.回想 上篇对 Volley进行了简介和对它的学习目的与目标,最后,为学习Volley做了一些准备 2.重点 2.1 RequestQueue 请求队列的建立 2.2 学习 StringReques ...

  3. RPC和Socket

    RPC和Socket的区别 rpc是通过什么实现啊?socket! RPC(Remote Procedure Call,远程过程调用)是建立在Socket之上的,出于一种类比的愿望,在一台机器上运行的 ...

  4. CentOS 与Ubuntu 安装软件包的对比

    工作需要开始转向centos,简单记录软件包安装 wget不是安装方式 他是一种下载软件类似与迅雷 如果要下载一个软件 我们可以直接 wget 下载地址 ap-get是ubuntu下的一个软件安装方式 ...

  5. C# MVC js 跨域

    js 跨域: 第一种解决方案(服务端解决跨域问题): 跨域是浏览器的一种安全策略,是浏览器自身做的限制,不允许用户访问不同域名或端口或协议的网站数据. 只有域名(主域名[一级域名]和二级域名).端口号 ...

  6. cf 828 A. Restaurant Tables

    A. Restaurant Tables time limit per test 1 second memory limit per test 256 megabytes input standard ...

  7. Fedora27 源配置

    一.添加阿里源,阿里源我感觉是现在国内比较好用的源,支持的发行版比较全.配置方法1.备份系统自带的源mv /etc/yum.repos.d/fedora.repo /etc/yum.repos.d/f ...

  8. MyEclipse的代码自动提示功能

     一般默认情况下,Eclipse ,MyEclipse的代码提示功能是比Microsoft Visual Studio的差很多的,主要是Eclipse ,MyEclipse本身有很多选项是默认关闭的, ...

  9. Linux下java/bin目录下的命令集合

    Linux下JAVA命令(1.7.0_79) 命令 详解 参数列表 示例 重要程度 资料 appletviewer Java applet 浏览器.appletviewer 命令可在脱离万维网浏览器环 ...

  10. Copying GC (Part one)

    目录 GC复制算法 copy()函数 将传递给自己的参数复制,然后递归复制其孩子 new_obj()函数 执行过程 缺点 Cheney的GC复制算法 copy()函数 执行过程 被隐藏的队列 优缺点 ...