zoukankan      html  css  js  c++  java
  • Codeforces Gym100952 A. Who is the winner? (2015 HIAST Collegiate Programming Contest)

    哈哈哈哈哈哈哈,最近一直在补题,改各种错误的代码,wa了20多遍,改到心态爆炸,改好之后,感觉世界都美好了(叉会腰~)。。。

     
    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

     

    代码:

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 int main(){
     4     int n;
     5     int a,b,c,a1,b1,c1;
     6     while(~scanf("%d",&n)){
     7         while(n--){
     8         scanf("%d%d%d%d%d%d",&a,&b,&c,&a1,&b1,&c1);
     9         if(a>a1) {printf("Player2
    ");}
    10         else if(a<a1) {printf("Player1
    ");}
    11         else if(a==a1){
    12             if(b>b1){printf("Player2
    ");}
    13             else if(b<b1){printf("Player1
    ");}
    14             else if(b==b1){
    15                 if(c>c1){printf("Player2
    ");}
    16                 else if(c<c1){printf("Player1
    ");}
    17                 else printf("Tie
    ");
    18             }
    19         }
    20         }
    21     }
    22     return 0;
    23 }
     
     
  • 相关阅读:
    图论集合
    无向连通图求割点(tarjan算法去掉改割点剩下的联通分量数目)
    河南省第七届ACM程序设计大赛总结
    单源最短路(spfa),删边求和
    最小圆覆盖
    二分图最大独立集
    二分图最少路径覆盖
    二分图最少点覆盖
    二分图最大匹配(匈牙利算法)
    最小费用最大流模板题
  • 原文地址:https://www.cnblogs.com/ZERO-/p/7245901.html
Copyright © 2011-2022 走看看