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

题目链接:http://codeforces.com/gym/100952/problem/A

题意:给定两组时间,包括时:分:秒三种信息,比较求解两组时间哪一个最短!
分析:直接做,忽略前导0,直接输入整数就好,当时以为是字符串输入,搞了半天都出不来!
下面给出AC代码:
 #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
inline int read()
{
int x=,f=;
char ch=getchar();
while(ch<''||ch>'')
{
if(ch=='-')
f=-;
ch=getchar();
}
while(ch>=''&&ch<='')
{
x=x*+ch-'';
ch=getchar();
}
return x*f;
}
inline void write(int x)
{
if(x<)
{
putchar('-');
x=-x;
}
if(x>)
{
write(x/);
}
putchar(x%+'');
}
int a[],b[];
int n;
int main()
{
n=read();
while(n--)
{
//cin.getline(s1,8);
//cin.getline(s2,8);
for(int i=;i<;i++)
a[i]=read();
for(int i=;i<;i++)
b[i]=read();
int flag=-;
for(int i=;i<;i++)
{
if(a[i]>b[i])
{
flag=;
break;
}
else if(a[i]<b[i])
{
flag=;
break;
}
else continue;
}
if(flag==)
printf("Player2\n");
else if(flag==)
printf("Player1\n");
else printf("Tie\n");
}
return ;
}

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

  4. 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 ...

  5. 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 ...

  6. 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 ...

  7. 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 ...

  8. 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 ...

  9. 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. web基础笔记整理(一)

    一.程序的分层 1.界面层: 某种类型的应用程序 a.DOS(控制台运行) b.桌面应用程序--独立安装,独立运行 c.web类型--现在流行的 单机版:电脑上要安装,程序升级之后,电脑上也要升级-- ...

  2. Hadoop版本选择

    刚开始学习Hadoop时就曾经一直抱怨Hadoop的安装部署为什么这么麻烦,对于一个新手需要捯饬一天才能把分布式环境安装配置好.而对于一个自学Hadoop而周围又没人交流的菜鸟来说,我对Hadoop的 ...

  3. sql server 2012 新知识-序列

    今天聊一聊sql 2012 上的新功能-----序列 按我的理解,它就是为了实现全局性的唯一标识,按sql server 以前的版本,想对一张表标识很简单,比如identity,但如果要对某几张有业务 ...

  4. RSA加解密实现

    RSA是由MIT的三位数学家R.L.Rivest,A.Shamir和L.Adleman[Rivest等1978, 1979]提出的一种用数论构造双钥的方法,被称为MIT体制,后来被广泛称之为RSA体制 ...

  5. 关于 dos 下 npm 命令的使用

    npm install 可以安装模块,后面跟 -g 安装全局的,后面跟包的名字就是安装指定的包 npm uninstall <安装包的名字> 卸载某个包,后面跟 -g 是卸载全局的某个包 ...

  6. 基于Dubbo的http自动测试工具分享

    公司是采用微服务来做模块化的,各个模块之间采用dubbo通信.好处就不用提了,省略了之前模块间复杂的http访问.不过也遇到一些问题: PS: Github的代码示例还在整理中... 测试需要配合写消 ...

  7. errcode 4103 invalid page hint 小程序模板消息推送遇到的坑

    invalid page hint一直提示这个坑爹的就是,我的小程序没发布之前,也就是测试版本用这个格式是可以的 /pages/myGroup/myGroup?groupid=22***但是发布成功以 ...

  8. Maven项目不打包*.hbm.xml文件

    <build> <finalName>basic</finalName> <plugins> <plugin> <groupId> ...

  9. 房上的猫:JavaDoc注释

    //这是一个注释 /*   *这是一个演示程序   */ /**    *@这是JavaDoc注释.   */ JavaDoc注释 背景: javadoc是Sun公司提供的一个技术,它从程序源代码中抽 ...

  10. git for windows上传项目到github

    软件:git for windows 账户:github账户 1.第一步创建自己的github账号,并创建自己的project,创建完毕之后url如下 https://github.com/ft110 ...