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
题意：输入n，表示n个人，第一个数表示做出了来的题数，第二个表示时间，做题数等于5得100分，4题时当做题时间排在前一半的时候得95，后一半得90，以此类推，题数等于0得0分。
/*放寒假以来第一次在没有空调的桌子旁写题，真的冷QAQ冷的没思路，看了大佬的题解才有了思路，代码也是模仿的
点击查看参（chao）考（xi）的代码QAQ */
我的代码
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct wzy{
int point,I,NO;
int h,m,s;
}p[200];
int cmp(wzy u,wzy v)
{
if(u.point==v.point)
{
if(u.h==v.h){
if(u.m==v.m){
return u.s<v.s;
}
else return u.m<v.m;
}
else return u.h<v.h;
}
return u.point>v.point;
}
int cmp1(wzy u,wzy v)
{
return u.I<v.I;
}
int main()
{
int n,i;
int k1,k2,k3,k4;
int b1,b2,b3,b4;
while(~scanf("%d",&n))
{
k1=k2=k3=k4=1;
b4=b3=b2=b1=0;
if(n<=0) break;
for(i=0;i<n;i++)
{
scanf("%d %d:%d:%d",&p[i].point,&p[i].h,&p[i].m,&p[i].s);
p[i].I=i;
if(p[i].point==4) b4++;
if(p[i].point==3) b3++;
if(p[i].point==2) b2++;
if(p[i].point==1) b1++;
}
sort(p,p+n,cmp);
for(i=0;i<n;i++)//记录每个题数所在的位置 （排名）
{
if(p[i].point==4) p[i].NO=k4++;
if(p[i].point==3) p[i].NO=k3++;
if(p[i].point==2) p[i].NO=k2++;
if(p[i].point==1) p[i].NO=k1++;
}
sort(p,p+n,cmp1);
for(i=0;i<n;i++)
{
if(p[i].point==5) printf("100\n");
if(p[i].point==4)
{
if(p[i].NO<=b4/2) printf("95\n");
else printf("90\n");
}
if(p[i].point==3)
{
if(p[i].NO<=b3/2) printf("85\n");
else printf("80\n");
}
if(p[i].point==2)
{
if(p[i].NO<=b2/2) printf("75\n");
else printf("70\n");
}
if(p[i].point==1)
{
if(p[i].NO<=b1/2) printf("65\n");
else printf("60\n");
}
if(p[i].point==0) printf("50\n");
}
printf("\n");
}
return 0;
}