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zsc_寒假训练 9

2024-11-12 22:39:02   203人阅读

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

45 06:30:174 07:31:274 08:12:124 05:23:1315 06:30:17-1
 

Sample Output

100909095100
 
技术分享
 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];//time13             int p;//problem14             int s;//score15     }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 }
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zsc_寒假训练 9


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