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; }