Description

“Point, point, life of student!”?
This is a ballad（歌謠）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（以此類推）, 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

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轉載于:https://www.cnblogs.com/guofeng1022/p/4241228.html

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