A. Who is the winner?
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
水题;
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
using namespace std;
typedef long long ll;
int main()
{
int a,b,c,x,y,z,n;
cin>>n;
while(n--)
{
cin>>a>>b>>c>>x>>y>>z;
int ans=a**+b*+c;
int pos=x**+y*+z;
if(ans>pos) puts("Player2");
else if(ans<pos) puts("Player1");
else puts("Tie");
}
}

Gym 100952 A. Who is the winner?的更多相关文章

  1. Gym 100952 D. Time to go back(杨辉三角形)

    D - Time to go back Gym - 100952D http://codeforces.com/gym/100952/problem/D D. Time to go back time ...

  2. codeforces gym 100952 A B C D E F G H I J

    gym 100952 A #include <iostream> #include<cstdio> #include<cmath> #include<cstr ...

  3. Gym 100952 H. Special Palindrome

    http://codeforces.com/gym/100952/problem/H H. Special Palindrome time limit per test 1 second memory ...

  4. Gym 100952 G. The jar of divisors

    http://codeforces.com/gym/100952/problem/G G. The jar of divisors time limit per test 2 seconds memo ...

  5. Gym 100952 F. Contestants Ranking

    http://codeforces.com/gym/100952/problem/F F. Contestants Ranking time limit per test 1 second memor ...

  6. Gym 100952 D. Time to go back

    http://codeforces.com/gym/100952/problem/D D. Time to go back time limit per test 1 second memory li ...

  7. Gym 100952 C. Palindrome Again !!

    http://codeforces.com/gym/100952/problem/C C. Palindrome Again !! time limit per test 1 second memor ...

  8. 【Gym 100712A】Who Is The Winner?

    题 题意 解题数目越多越排前,解题数目相同罚时越少越排前,求排第一的队伍名字. 分析 用结构体排序. 代码 #include<cstdio> #include<algorithm&g ...

  9. Gym 100952 B. New Job

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

随机推荐

  1. CDH5.14操作指南

    1.Cdh 5.14介绍 2.版本发布概要 3.安装要求 4.Cloudera Manager介绍 5.Cloudera 安装 6.Cloudera 管理 7.Cloudera Navigator C ...

  2. 洛谷P3332 [ZJOI2013]K大数查询 权值线段树套区间线段树_标记永久化

    Code: #include <cstdio> #include <algorithm> #include <string> #include <cstrin ...

  3. NodeJS学习笔记 进阶 (1)Nodejs进阶:服务端字符编解码&乱码处理(ok)

    个人总结:这篇文章主要讲解了Nodejs处理服务器乱码及编码的知识,读完这篇文章需要10分钟. 摘选自网络 写在前面 在web服务端开发中,字符的编解码几乎每天都要打交道.编解码一旦处理不当,就会出现 ...

  4. 移动和PC的适配

    <script> //mode 移动端的适配方式 按需 传参 目前只有两种 px和rem (function(win, doc, mode) { var std = 750; if(/(i ...

  5. bzoj2100 [Usaco2010 DEC]Apple Delivery苹果贸易

    题目描述 一张P个点的无向图,C条正权路.CLJ要从Pb点(家)出发,既要去Pa1点NOI赛场拿金牌,也要去Pa2点CMO赛场拿金牌.(途中不必回家)可以先去NOI,也可以先去CMO.当然神犇CLJ肯 ...

  6. mysql(for update)悲观锁总结与实践

    悲观锁,正如其名,它指的是对数据被外界(包括本系统当前的其他事务,以及来自外部系统的事务处理)修改持保守态度,因此,在整个数据处理过程中,将数据处于锁定状态.悲观锁的实现,往往依靠数据库提供的锁机制( ...

  7. django-xadmin使用之更改菜单url

    环境:xadmin-for-python3 python3.5.2 django1.9.12 1. 在模块的adminx.py文件中增加以下代码: class AdminSettings(object ...

  8. 自己定义View之Chart图标系列(1)——点阵图

    近期要做一些图表类的需求,一開始就去github上看了看,发现开源的图表框架还是蛮多的.可是非常少有全然符合我的需求的.另外就是使用起来比較麻烦.所以就决定自己来造轮子了~~~ 今天要介绍的就是And ...

  9. Win8.1应用开发之文件操作

    在操作文件之前,先相应用的应用功能声明进行设定.用户通过C#(非UI)对win8.1上的文件进行訪问,仅仅能局限于图片,音乐,视频和文档四个目录. 而通过文件选取器则能訪问到整个系统的文件. (一)应 ...

  10. (iOS)确保设置话筒模式成功 AudioSessionSetProperty

    本人编写过一个应用,须要把实时音频播放出来,而且要从话筒播放声音,为此,作下面操作: //Step 1: 初始化 AudioSessionInitialize(NULL,NULL, NULL, sel ...