十二、What Is Your Grade?

Problem 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

输入一组即输出结果；再输入下一组数据，以此类推；

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>using namespace std;struct T{    int begin;    int acomplish_work;    int consume_time;    int score;} t[100];int cmp(T a,T b){    if(a.acomplish_work>b.acomplish_work) return 1;     else if(a.acomplish_work==b.acomplish_work) return a.consume_time<b.consume_time;    else return 0;}int cmp1(T a,T b){      return a.begin<b.begin;}int main(){    int group;    int i,j,k;    char c;    int a[5];    int x,y,z;    while((cin>>group)&&(group>=0))    {    memset(a,0,sizeof(a));    for(i=0;i<group;i++)    {        t[i].begin=i;//原始的名次排列        t[i].consume_time=0;        cin>>t[i].acomplish_work; //每个人完成的作业量        a[t[i].acomplish_work]++;//累计每个名次的人数        t[i].score = 100 - (5 - t[i].acomplish_work) * 10;      cin>>x>>c>>y>>c>>z;       t[i].consume_time=x*3600+y*60+z;//每一个人的总耗时}     sort(t,t+group,cmp);//按照完成题目的数量和数量相同的按时间消耗较小的进行排序     for(i = 4, j = 0; i; --i){//参考代码                 if(a[i]){                     while(t[j].acomplish_work != i) ++j;                     if(a[i] == 1) t[j++].score += 5;                     for( k = 0; k < a[i] / 2; ++k)                         t[j++].score += 5;                 }             }    sort(t,t+group,cmp1);//按照序号进行正序    for(i=0;i<group;i++)//   cout<<t[i].begin<<"  "<<t[i].acomplish_work<<"  "<<t[i].consume_time<<" "<<t[i].score<<endl;     cout<<t[i].score<<endl;    cout<<endl;    }    return 0;}