He is offside!

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 538 Accepted Submission(s):
333

Problem Description
Hemisphere Network is the largest television network in
Tumbolia, a small country located east of South America (or south of East
America). The most popular sport in Tumbolia, unsurprisingly, is soccer; many
games are broadcast every week in Tumbolia.

Hemisphere Network receives
many requests to replay dubious plays; usually, these happen when a player is
deemed to be offside by the referee. An attacking player is offside if he is
nearer to his opponents' goal line than the second last opponent. A player is
not offside if
●he is level with the second last opponent or
●he is level
with the last two opponents.

Through the use of computer graphics
technology, Hemisphere Network can take an image of the field and determine the
distances of the players to the defending team's goal line, but they still need
a program that, given these distances, decides whether a player is
offside.

 
Input
The input file contains several test cases. The first
line of each test case contains two integers A and D separated by a single space
indicating, respectively, the number of attacking and defending players involved
in the play (2 <= A,D <= 11). The next line contains A integers Bi
separated by single spaces, indicating the distances of the attacking players to
the goal line
(1 <= Bi <= 104). The next line contains D
integers Cj separated by single spaces, indicating the distances of the
defending players to the goal line (1 <= Cj <= 104). The end of
input is indicated by A = D = 0.
 
Output
For each test case in the input print a line containing
a single character: "Y"(uppercase) if there is an attacking player offside, and
"N"(uppercase) otherwise.
 
Sample Input
2 3
500 700
700 500 500
2 2
200 400
200 1000
3 4
530 510 490
480 470 50 310
0 0
 
Sample Output
N
Y
N
 
Source
 
 
 
 
 
 
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int t1[],t2[];
int main(){
int a,b;
while(scanf("%d%d",&a,&b)!=EOF){
if(a==&&b==)
break;
memset(t1,,sizeof(t1));
memset(t2,,sizeof(t2));
for(int i=;i<a;i++)
scanf("%d",&t1[i]);
for(int i=;i<b;i++)
scanf("%d",&t2[i]);
sort(t1,t1+a);
sort(t2,t2+b);
if(t1[]<t2[])
printf("Y\n");
else
printf("N\n"); } return ;
}

HDU 1939 HE IS OFFSIDE的更多相关文章

  1. hdu 3415 Max Sum of Max-K-sub-sequence 单调队列。

    Max Sum of Max-K-sub-sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K ...

  2. HDU 5643 King's Game 打表

    King's Game 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5643 Description In order to remember hi ...

  3. 转载:hdu 题目分类 (侵删)

    转载:from http://blog.csdn.net/qq_28236309/article/details/47818349 基础题:1000.1001.1004.1005.1008.1012. ...

  4. HDOJ 2111. Saving HDU 贪心 结构体排序

    Saving HDU Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total ...

  5. 【HDU 3037】Saving Beans Lucas定理模板

    http://acm.hdu.edu.cn/showproblem.php?pid=3037 Lucas定理模板. 现在才写,noip滚粗前兆QAQ #include<cstdio> #i ...

  6. hdu 4859 海岸线 Bestcoder Round 1

    http://acm.hdu.edu.cn/showproblem.php?pid=4859 题目大意: 在一个矩形周围都是海,这个矩形中有陆地,深海和浅海.浅海是可以填成陆地的. 求最多有多少条方格 ...

  7. HDU 4569 Special equations(取模)

    Special equations Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u S ...

  8. HDU 4006The kth great number(K大数 +小顶堆)

    The kth great number Time Limit:1000MS     Memory Limit:65768KB     64bit IO Format:%I64d & %I64 ...

  9. HDU 1796How many integers can you find(容斥原理)

    How many integers can you find Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%I64d ...

随机推荐

  1. mysql关键字了解

    unsigned  无符号 就是没有负数 列-1  -2 auto_increment 自增 comment 注释 primary key 主键 foreign key ()   references ...

  2. 第26章 FMC—扩展外部SDRAM—零死角玩转STM32-F429系列

    第26章     FMC—扩展外部SDRAM 全套200集视频教程和1000页PDF教程请到秉火论坛下载:www.firebbs.cn 野火视频教程优酷观看网址:http://i.youku.com/ ...

  3. 关于event loop

    之前写了篇文章 JS运行机制,里面对event loop简单的说明,面试时又遇到了关于该知识点的题目(主要是process.nextTick和setImmediate的执行顺序不太知道,查了之后才知道 ...

  4. C++声明之CV限定符

    目录 1.const 1.1 const obj 如果调用 non-const member fun会编译出错 经典错误 1.2 例子:STD里的操作符重载 1.3 例子:<cpp primer ...

  5. hibernate系列之四

    数据库中表之间的关系: 一对一.一对多.多对多 一对多的建表原则:在多的一方创建外键指向一的一方的主键: 多对多的建表原则:创建一个中间表,中间表中至少有两个字段作为外键分别指向多对多双方的主键: 一 ...

  6. yii rbac

    一.简介 什么是rbac ? rbac是就是基于角色的访问控制. yii提供一套基础的底层接口,我们知道,rbac经历好几个阶段,从rbac0到rbac3,从基础的用户.角色.权限,到动态的rbac处 ...

  7. C指针——简单总结

    简介: 指针变量在使用前,必须指向具体的有效的内存单元 指针变量在使用前不但要定义还要初始化 四个方面:指针的类型,指针指向的类型,指针的值或者指针所指向的内存区,指针本身所占的内存区 int *pt ...

  8. python-7面向对象高级编程

    1-给类动态增加方法 class Student(object): pass def set_score(self, score): self.score = score Student.set_sc ...

  9. Eclipse字体修改

    第一步: 第二步: 第三步: 第四步: 第五步: 第六步:

  10. 如何使用Python脚本

    来自官方文档 一.写 python 脚本: import sys import datetime for line in sys.stdin: line = line.strip() userid, ...