HDU 2115 I Love This Game

http://acm.hdu.edu.cn/showproblem.php?pid=2115

 

I Love This Game

TimeLimit: 3000/1000 MS(Java/Others)    MemoryLimit: 32768/32768 K (Java/Others)
Total Submission(s):2968    AcceptedSubmission(s): 1026

Problem Description

Do you like playing basketball ? If youare , you may know the NBA Skills Challenge . It is the content ofthe basketball skills . It include several parts , such as passing, shooting , and so on. After completion of the content , theplayer who takes the shortest time will be the winner . Now giveyou their names and the time of finishing the competition , yourtask is to give out the rank of them ; please output their name andthe rank, if they have the same time , the rank of them will be thesame ,but you should output their names in lexicographic order.Youmay assume the names of the players are unique.

Is it a very simple problem for you? Please accept it in tenminutes.

 

Input

This problem contains multiple testcases! Ease test case contain an(1<=n<=10) shows the number ofplayers,then n lines will be given. Each line will contain the nameof player and the time(mm:ss) of their finish.The end of the inputwill be indicated by an integer value of zero.

 

Output

The output format is shown as samplebelow.
Please output the rank of all players, the output format is shownas sample below;
Output a blank line between two cases.

 

Sample Input

10 Iverson17:19 Bryant 07:03 Nash 09:33 Wade 07:03 Davies 11:13 Carter 14:28Jordan 29:34 James 20:48 Parker 24:49 Kidd 26:46 0

 

Sample Output

Case #1Bryant 1 Wade 1 Nash 3 Davies 4 Carter 5 Iverson 6 James 7 Parker 8Kidd 9 Jordan 10

 

Author

為傑沉倫

 

Source

HDU 2007-10 Programming Contest_WarmUp

 

Recommend

威士忌

 

题意：按照给出的时间排序。有两点注意：每两个输出之间空一行、排名会出现并列。

分析：用结构体，把信息捆绑在一起，记录时间，排序输出即可。这个题的排序，我建议用冒泡排序，因为qsort还得写cmp函数。我的代码0ms过的，从网上找了一个用qsort的，15ms^-^

代码如下：

#include<stdio.h>
struct pp
{
 char w[20];
 int shi;
 int fen;
 int shijian;
 int paiming;
};
pp ans[11];
pp temp;
int main()
{
 int n,i,j,m=1;
 while(scanf("%d",&n)!=EOF&&n!=0)
 {
    for(i=0;i<n;i++)
   {
   scanf("%s%d:%d",ans[i].w,&ans[i].shi,&ans[i].fen);
   }
    for(i=0;i<n;i++)
    ans[i].shijian=60*ans[i].shi+ans[i].fen;
   for(i=0;i<n;i++)
     {
      for(j=0;j<n-1-i;j++)          //冒泡排序好久不用了，j的for循环条件写错了，wrong个无数次。。
         {
   if(ans[j].shijian>ans[j+1].shijian)
           {
            temp=ans[j];
            ans[j]=ans[j+1];
            ans[j+1]=temp;
           }
         }
     }
     ans[0].paiming=1;
     for(i=1;i<n;i++)
       {
        if(ans[i-1].shijian==ans[i].shijian)ans[i].paiming=ans[i-1].paiming;
        else ans[i].paiming=i+1;
       }
       if(m!=1)
  printf("\n");
       printf("Case #%d\n",m++);
       for(i=0;i<n;i++)
           printf("%s %d\n",ans[i].w,ans[i].paiming);
 }
 return 0;
}