“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
解题思路：这道题就是使用sort+结构体就可以了
AC代码：
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=110;
struct node{
char time[20];
int num;//解题数
int id;//输入学生的顺序
int rank;//学生在当前解题数的排名，用时越短排名越靠前，
int score;//学生的成绩
}a[maxn];
bool cmp(node a,node b)//按解题数从大到小排序，解题数相同时，按用时从小到大排序，
{
if(a.num==b.num)
return strcmp(a.time,b.time)<0;
return a.num>b.num;
}
bool cmp1(node a,node b)//按学生输入顺序排序
{
return a.id<b.id;
}
int main()
{
int n;
while(cin>>n)
{
if(n<=0)
break;
int c[6]={0};//储存解题数为n时的人数n= 0 1 2 3 4 5,初始化为0
for(int i=0;i<n;i++)
{
cin>>a[i].num>>a[i].time;
a[i].id=i;
}
sort(a,a+n,cmp);
for(int i=0;i<n;i++)
{
a[i].rank=++c[a[i].num];//排名当前解题数为a[i].num学生中的排名，
}
for(int i=0;i<6;i++)
{
c[i]/=2;//计算解题数为i的前一半的人数为多少
}
for(int i=0;i<n;i++)//计算成绩
{
if(a[i].num==5)
a[i].score=100;
else if(a[i].num==4)
{
if(a[i].rank<=c[a[i].num])
a[i].score=95;
else
a[i].score=90;
}
else if(a[i].num==3)
{
if(a[i].rank<=c[a[i].num])
a[i].score=85;
else
a[i].score=80;
}
else if(a[i].num==2)
{
if(a[i].rank<=c[a[i].num])
a[i].score=75;
else
a[i].score=70;
}
else if(a[i].num==1)
{
if(a[i].rank<=c[a[i].num])
a[i].score=65;
else
a[i].score=60;
}
else
a[i].score=50;
}
sort(a,a+n,cmp1);
for(int i=0;i<n;i++)
{
cout<<a[i].score<<endl;
}
cout<<endl;
}
return 0;
}