time limit per test

1 second

memory limit per test

64 megabytes

input

standard input

output

standard output

A big marathon is held on Al-Maza Road, Damascus. Runners came from all over the world to run all the way along the road in this big marathon day. The winner is the player who crosses the finish line first.

The organizers write down finish line crossing time for each player. After the end of the marathon, they had a doubt between 2 possible winners named "Player1" and "Player2". They will give you the crossing time for those players and they want you to say who is the winner?

Input

First line contains number of test cases (1  ≤  T  ≤  100). Each of the next T lines represents one test case with 6 integers H1 M1 S1 H2 M2 S2. Where H1, M1, S1 represent Player1 crossing time (hours, minutes, seconds) and H2, M2, S2 represent Player2 crossing time (hours, minutes, seconds). You can assume that Player1 and Player2 crossing times are on the same day and they are represented in 24 hours format(0  ≤  H1,H2  ≤  23 and 0  ≤  M1,M2,S1,S2  ≤  59)

H1, M1, S1, H2, M2 and S2 will be represented by exactly 2 digits (leading zeros if necessary).

Output

For each test case, you should print one line containing "Player1" if Player1 is the winner, "Player2" if Player2 is the winner or "Tie" if there is a tie.

Examples
Input
3
18 03 04 14 03 05
09 45 33 12 03 01
06 36 03 06 36 03
Output
Player2
Player1
Tie
 

题意就是比较谁花的时间少,水题

代码:

#include<bits/stdc++.h>
using namespace std;
int main(){
int n;
int a,b,c,a1,b1,c1;
while(~scanf("%d",&n)){
while(n--){
scanf("%d%d%d%d%d%d",&a,&b,&c,&a1,&b1,&c1);
if(a>a1) {printf("Player2\n");}
else if(a<a1) {printf("Player1\n");}
else if(a==a1){
if(b>b1){printf("Player2\n");}
else if(b<b1){printf("Player1\n");}
else if(b==b1){
if(c>c1){printf("Player2\n");}
else if(c<c1){printf("Player1\n");}
else printf("Tie\n");
}
}
}
}
return ;
}

CodeForces-2015 HIAST Collegiate Programming Contest-Gym-100952A-Who is the winner?的更多相关文章

  1. Gym 100952E&&2015 HIAST Collegiate Programming Contest E. Arrange Teams【DFS+剪枝】

    E. Arrange Teams time limit per test:2 seconds memory limit per test:64 megabytes input:standard inp ...

  2. Gym 100952F&&2015 HIAST Collegiate Programming Contest F. Contestants Ranking【BFS+STL乱搞(map+vector)+优先队列】

    F. Contestants Ranking time limit per test:1 second memory limit per test:24 megabytes input:standar ...

  3. Gym 100952A&&2015 HIAST Collegiate Programming Contest A. Who is the winner?【字符串,暴力】

    A. Who is the winner? time limit per test:1 second memory limit per test:64 megabytes input:standard ...

  4. Gym 100952J&&2015 HIAST Collegiate Programming Contest J. Polygons Intersection【计算几何求解两个凸多边形的相交面积板子题】

    J. Polygons Intersection time limit per test:2 seconds memory limit per test:64 megabytes input:stan ...

  5. Gym 100952I&&2015 HIAST Collegiate Programming Contest I. Mancala【模拟】

    I. Mancala time limit per test:3 seconds memory limit per test:256 megabytes input:standard input ou ...

  6. Gym 100952H&&2015 HIAST Collegiate Programming Contest H. Special Palindrome【dp预处理+矩阵快速幂/打表解法】

    H. Special Palindrome time limit per test:1 second memory limit per test:64 megabytes input:standard ...

  7. Gym 100952G&&2015 HIAST Collegiate Programming Contest G. The jar of divisors【简单博弈】

    G. The jar of divisors time limit per test:2 seconds memory limit per test:64 megabytes input:standa ...

  8. Gym 100952D&&2015 HIAST Collegiate Programming Contest D. Time to go back【杨辉三角预处理,组合数,dp】

    D. Time to go back time limit per test:1 second memory limit per test:256 megabytes input:standard i ...

  9. Gym 100952C&&2015 HIAST Collegiate Programming Contest C. Palindrome Again !!【字符串,模拟】

    C. Palindrome Again !! time limit per test:1 second memory limit per test:64 megabytes input:standar ...

  10. Gym 100952B&&2015 HIAST Collegiate Programming Contest B. New Job【模拟】

    B. New Job time limit per test:1 second memory limit per test:64 megabytes input:standard input outp ...

随机推荐

  1. getchar() 、 scanf() 、流与缓冲区

    C中的缓冲区一直是debug的重灾区,今天在写一个命令行界面的时候又遇到了这个问题,所以来总结一波. 两函数的不同之处 scanf() 会把 stdinBuff 中的特定格式数据取出,非特定格式数据则 ...

  2. Python学习日记:day9--------函数

    初识函数 1,自定义函数 s ='内容' #自定义函数 def my_len():#自定义函数没有参数 i =0 for k in s: i+=1 print(i) return i #返回值 my_ ...

  3. 第四节:dingo/API 最新版 V2.0 之 Responses (连载)

    因为某些某些原因,不能按时更新,唉.我会尽力,加快速度.(这句话不是翻译的哈) 原文地址--> https://github.com/dingo/api/wiki/Responses A fun ...

  4. ArcGIS 网络分析[2] 在ArcMap中使用网络数据集进行五大网络分析[最短路径/服务区/最近设施点/OD成本矩阵/车辆分配]

    上一章花了大篇幅介绍网络数据集的创建,也简单说了下点线的连通性问题. 那么可以试试刀锋不锋利啦! 网络分析呢,ArcGIS提供了5个基本分析类型: 最短路径求解 服务区(服务覆盖范围) 事故突发地的最 ...

  5. 初学者福音——10个最佳APP开发入门在线学习网站

    根据Payscale的调查显示,现在的APP开发人员的年薪达到:$66,851.这也是为什么那么多初学的开发都想跻身到APP开发这行业的主要原因之一.每当你打开App Store时候,看着琳琅满目的A ...

  6. Oracle学习笔记_09_字符串相关函数

    二.参考资料 0.Oracle中的字符串类型及相关函数详解 1.ORACLE 字符串操作 2.oracle函数大全-字符串处理函数

  7. Sublime Text 3 配置分析与我的配置---小结

    Sublime Text 3 配置解释(默认){// 设置主题文件"color_scheme": "Packages/Color Scheme – Default/Mon ...

  8. [编织消息框架][netty源码分析]14 PoolChunk 的 PoolSubpage

    final class PoolSubpage<T> implements PoolSubpageMetric { //该page分配的chunk final PoolChunk<T ...

  9. 3D轮播切换特效 源码

    这个3D轮播切换特效是我2017年2月份写的 当初我 刚接触HTML不久,现在把源码分享给大家 源码的注释超级清楚 . <!-- 声明文档类型:html 作用:符合w3c统一标准规范 每个浏览器 ...

  10. CentOS7源码安装lamp

    环境介绍 虚拟机 : VMware Workstation 14 Pro 镜像 : CentOS Linux release 7.4.1708 (Core) 物理机 : windows 7 64位 防 ...