CodeForces-2015 HIAST Collegiate Programming Contest-Gym-100952A-Who is the winner?
1 second
64 megabytes
standard input
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?
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).
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.
3
18 03 04 14 03 05
09 45 33 12 03 01
06 36 03 06 36 03
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
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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
- 转:Siri之父:语音交互或将主导未来十年发展
http://zhinengjiaohu.juhangye.com/201709/weixin_5664458.html Siri之父Adam Cheyer认为,语音交互很可能是未来十年内计算技术的一 ...
- Java 银行家算法
实验存档,代码特别烂.. 测试.java package operating.test; import operating.entity.bank.Bank; import operating.ent ...
- xamarin android menu的用法
在Android中的菜单有如下几种: OptionMenu:选项菜单,android中最常见的菜单,通过Menu键来调用 SubMenu:子菜单,android中点击子菜单将弹出一个显示子菜单项的悬浮 ...
- ArcGIS API for JavaScript 4.2学习笔记[12] View的弹窗(Popup)
看本文前最好对第二章(Mapping and Views)中的Map和View类有理解. 视图类有一个属性是Popup类型的popup,查阅API知道这个就是视图的弹窗,每一个View的实例都有一个p ...
- headfirst设计模式(4)—工厂模式
开篇 天天逛博客园,就是狠不下心来写篇博客,忙是一方面,但是说忙能有多忙呢,都有时间逛博客园,写篇博客的时间都没有?(这还真不好说) 每次想到写一篇新的设计模式,我总会问自己: 1,自己理解了吗? 2 ...
- HDFS租约实践
一.租约详解 Why租约 HDFS的读写模式为 "write-once-read-many",为了实现write-once,需要设计一种互斥机制,租约应运而生租约本质上是一个有时间 ...
- 模板引擎(smarty)知识点总结II
今天咱们继续来学习smarty!!! 知识点1:对于三种变量 常量的引用 有哪三种变量?a.assign赋值 b.系统保留变量(包括:$smarty.get,$smarty.post,$smarty. ...
- Java自己动手写连接池四
Java自己动手写连接池四 测试: package com.kama.cn; import java.sql.Connection; public class Test { public static ...
- 鸟哥的linux私房菜学习-(一)优缺点分析以及主机规划与磁盘分区
一.linux的优缺点 那干嘛要使用Linux做为我们的主机系统呢?这是因为Linux有底下这些优点: 稳定的系统:Linux本来就是基于Unix概念而发展出来的操作系统,因此,Linux具有与Uni ...
- JavaMail开发教程01开山篇
序 其实想写JavaMail这一系列的博客已经有一个月之久了,缘起是某次乱逛传智播客官网浏览到相关的视频教程,想起大学时代学过的计算机网络提到邮件相关的协议,但遗憾的是到目前为止还没有接触计算机网络编 ...