“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
关键点：做对1~4题的人的时间从小到大排序，前半部分分别给65,75,85,95分，另外，需要记录学生的位置，方便排序后恢复成原来的序列。
Code： ``` #include<iostream> #include<algorithm> #include<cstdio> using namespace std; struct node { int tm; int sum; int id; int same;//重复人数 int score; }; bool cmp1(node x,node y){ if(x.tm>y.tm) return 1; else if(x.tm==y.tm){ if(x.sum<y.sum) return 1; return 0; } else return 0; } bool cmp2(node x,node y){ if(x.id<y.id) return 1; else return 0; } int main(){ int n; while(cin>>n&&n>=0){ node a[n]; int s1=0,s2=0,s3=0,s4=0; for(int i=0;i<n;i++){ int x,y,z;//时分秒 scanf("%d %d:%d:%d",&a[i].tm,&x,&y,&z); if(a[i].tm==1) s1++; if(a[i].tm==2) s2++; if(a[i].tm==3) s3++; if(a[i].tm==4) s4++; a[i].sum=x*3600+y*60+z; a[i].id=i; } sort(a,a+n,cmp1); a[0].same=0; for(int i=1;i<n;i++){ if(a[i].tm==a[i-1].tm){ a[i].same=a[i-1].same+1; } else a[i].same=0; } for(int i=0;i<n;i++){ switch(a[i].tm){ case 1:if(a[i].same<s1/2) a[i].score=65; else a[i].score=60;break; case 2:if(a[i].same<s2/2) a[i].score=75; else a[i].score=70;break; case 3:if(a[i].same<s3/2) a[i].score=85; else a[i].score=80;break; case 4:if(a[i].same<s4/2) a[i].score=95; else a[i].score=90;break; case 5:a[i].score=100;break; default :a[i].score=50; } } sort(a,a+n,cmp2); for(int i=0;i<n;i++){ cout<<a[i].score<<endl; } cout<<endl; } return 0; } ```