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. UI Framework-1: Aura Views

    Views Views is a user interface framework built on a type called, confusingly, View. Responsible for ...

  2. dijkstra STL 堆优化

    Code: #include<iostream> #include<algorithm> #include<vector> #include<queue> ...

  3. 去除input的前后的空格

    这里用的是jquery的方法

  4. [NOIP2009提高组]靶形数独

    题目:洛谷P1074.Vijos P1755.codevs1174. 题目大意:给你一个数独,让你填完这个数独,并要求得分最大,问这个得分是多少(不能填完输出-1). 每个格子的得分是当前格子所填的数 ...

  5. dmesg---检查和控制内核的环形缓冲区

    dmesg命令被用于检查和控制内核的环形缓冲区.kernel会将开机信息存储在ring buffer中.您若是开机时来不及查看信息,可利用dmesg来查看.开机信息保存在/var/log/dmesg文 ...

  6. 使用maven安装jar到本地仓库

    mvn install:install-file "-DgroupId={安装的jar包的groupid,可以随意起名}" "-DartifactId={安装jar包的I ...

  7. Android如何从外部跳进App

    博客出自:http://blog.csdn.net/liuxian13183,转载注明出处! All Rights Reserved ! 这个问题解决了两天时间,因为网上没有完整的解决方案,解决后分享 ...

  8. 三 概要模式 2) MR倒排索引、性能分析、搜索干扰词。

    二  倒排索引     倒排索引(英语:Inverted index),也常被称为反向索引.置入档案或反向档案,是一种索引方法,被用来存储在全文搜索下某个单词在一个文档或者一组文档中的存储位置的映射. ...

  9. 【VBA研究】用VBA取得EXCEL随意列有效行数

    作者:iamlaosong 用VBA对Excel文件进行处理的时候,keyword段的列号编程时往往是不知道的.须要通过參数设定才干知道,因此.我们编程的时候,就不能用这种语句取有效行数: linen ...

  10. CAShapeLayer的简单介绍以及基本使用

    1.CAShapeLayer简单介绍  1.1CAShapeLayer继承于CALayer,能够使用CALayer的全部属性值:    1.2CAShapeLayer须要贝塞尔曲线配合使用才有意义(也 ...