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
318 03 04 14 03 0509 45 33 12 03 0106 36 03 06 36 03
Output
Player2Player1Tie
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 0;
}
。。。