CDZSC_2015寒假新人（1）——基础 i

Description

“Point, point, life of student!” 
This is a ballad（歌谣）well known in colleges, and you must care about your score in this exam too. How many points can you get? Now, I told you the rules which are used in this course. 
There are 5 problems in this final exam. And I will give you 100 points if you can solve all 5 problems; of course, it is fairly difficulty for many of you. If you can solve 4 problems, you can also get a high score 95 or 90 (you can get the former(前者) only when your rank is in the first half of all students who solve 4 problems). Analogically（以此类推）, you can get 85、80、75、70、65、60. But you will not pass this exam if you solve nothing problem, and I will mark your score with 50. 
Note, only 1 student will get the score 95 when 3 students have solved 4 problems. 
I wish you all can pass the exam! 
Come on! 

 

Input

Input contains multiple test cases. Each test case contains an integer N (1<=N<=100, the number of students) in a line first, and then N lines follow. Each line contains P (0<=P<=5 number of problems that have been solved) and T（consumed time）. You can assume that all data are different when 0<p. 
A test case starting with a negative integer terminates the input and this test case should not to be processed. 

 

Output

Output the scores of N students in N lines for each case, and there is a blank line after each case. 

 

Sample Input

4 5 06:30:17 4 07:31:27 4 08:12:12 4 05:23:13 1 5 06:30:17 -1

 

Sample Output

100 90 90 95 100

 

 1 #include<iostream>

 2 #include<cstring>

 3 #include<cstdio>

 4 #include<algorithm>

 5 using namespace std;

 6 

 7 int main()

 8 {

 9     int n;

10     struct 

11     {

12             char t[10];//time

13             int p;//problem

14             int s;//score

15     }a[120];

16     while((scanf("%d",&n))&&n!=-1)

17     {

18         for(int i=0;i<n;i++)

19         {

20             scanf("%d%s",&a[i].p,&a[i].t);

21             a[i].s=50+10*a[i].p;

22         }

23         int x=1;

24         while(x<5)

25         {

26             char time[120][10]={"99:99:99"};

27             int num=0;

28             

29             for(int i=0;i<n;i++)

30             {

31                 if(a[i].p==x)

32                 {

33                     strcpy(time[num++],a[i].t);

34                 }

35             }

36             for(int i=0;i<num/2;i++)

37             {

38                 for(int j=i+1;j<num;j++)

39                 {

40                     if(strcmp(time[i],time[j])>0)

41                     {

42                         char p[10];

43                         strcpy(p,time[i]);

44                         strcpy(time[i],time[j]);

45                         strcpy(time[j],p);

46                     }

47                 }

48             }

49             for(int i=0;i<n;i++)

50             {

51                 if(a[i].p==x&&strcmp(a[i].t,time[num/2-1])<=0)

52                 {

53                     a[i].s+=5;

54                 }

55             }

56             x++;

57         }

58         for(int i=0;i<n;i++)

59         {

60             printf("%d\n",a[i].s);

61         }

62         printf("\n");

63     }

64 }

View Code