Gym 100952A&&2015 HIAST Collegiate Programming Contest A. Who is the winner?【字符串,暴力】
A. Who is the winner?
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
题目链接:http://codeforces.com/gym/100952/problem/A
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 inline int read() 5 { 6 int x=0,f=1; 7 char ch=getchar(); 8 while(ch<'0'||ch>'9') 9 { 10 if(ch=='-') 11 f=-1; 12 ch=getchar(); 13 } 14 while(ch>='0'&&ch<='9') 15 { 16 x=x*10+ch-'0'; 17 ch=getchar(); 18 } 19 return x*f; 20 } 21 inline void write(int x) 22 { 23 if(x<0) 24 { 25 putchar('-'); 26 x=-x; 27 } 28 if(x>9) 29 { 30 write(x/10); 31 } 32 putchar(x%10+'0'); 33 } 34 int a[3],b[3]; 35 int n; 36 int main() 37 { 38 n=read(); 39 while(n--) 40 { 41 //cin.getline(s1,8); 42 //cin.getline(s2,8); 43 for(int i=0;i<3;i++) 44 a[i]=read(); 45 for(int i=0;i<3;i++) 46 b[i]=read(); 47 int flag=-1; 48 for(int i=0;i<3;i++) 49 { 50 if(a[i]>b[i]) 51 { 52 flag=1; 53 break; 54 } 55 else if(a[i]<b[i]) 56 { 57 flag=0; 58 break; 59 } 60 else continue; 61 } 62 if(flag==1) 63 printf("Player2\n"); 64 else if(flag==0) 65 printf("Player1\n"); 66 else printf("Tie\n"); 67 } 68 return 0; 69 }
作 者:Angel_Kitty
出 处:https://www.cnblogs.com/ECJTUACM-873284962/
关于作者:阿里云ACE,目前主要研究方向是Web安全漏洞以及反序列化。如有问题或建议,请多多赐教!
版权声明:本文版权归作者和博客园共有,欢迎转载,但未经作者同意必须保留此段声明,且在文章页面明显位置给出原文链接。
特此声明:所有评论和私信都会在第一时间回复。也欢迎园子的大大们指正错误,共同进步。或者直接私信我
声援博主:如果您觉得文章对您有帮助,可以点击文章右下角【推荐】一下。您的鼓励是作者坚持原创和持续写作的最大动力!
欢迎大家关注我的微信公众号IT老实人(IThonest),如果您觉得文章对您有很大的帮助,您可以考虑赏博主一杯咖啡以资鼓励,您的肯定将是我最大的动力。thx.
我的公众号是IT老实人(IThonest),一个有故事的公众号,欢迎大家来这里讨论,共同进步,不断学习才能不断进步。扫下面的二维码或者收藏下面的二维码关注吧(长按下面的二维码图片、并选择识别图中的二维码),个人QQ和微信的二维码也已给出,扫描下面👇的二维码一起来讨论吧!!!
欢迎大家关注我的Github,一些文章的备份和平常做的一些项目会存放在这里。