F

“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! 

InputInput 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. 
OutputOutput 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

刚看到这个题目的时候有点懵，但是尝试着写了一下。。。。思路简单的很

#include<iostream>
#include<algorithm>
#include<string> 
#include<cstdio>
#include<math.h>
using namespace std;
struct stu
{
    int a;
    string b;
    int c;//记录当前的分数 
};
//先用sort对副本数组num2排序，a优先，b其次，c不用管 ，最后在转换为num1
bool cmp(stu x,stu y)
{
    if(x.a!=y.a)
        return x.a>y.a;
    return x.b<y.b;//时间取较小的 
}
int main()
{
    int n;
    while(cin>>n)
    {
        int s[6]={0};//，一定要定义在循环内部，，，错了好几次 
        struct stu num1[111],num2[111];
        if(n<0)
            break;
            
        for(int i=0;i<n;i++)
        {
            cin>>num1[i].a>>num1[i].b;
            num2[i].a=num1[i].a;
            num2[i].b=num1[i].b;
            s[num1[i].a]++;//记录每个值出现的次数 
        }
        
        sort(num2,num2+n,cmp);
        int x=0,y=0,z=0,w=0;
        for(int i=0;i<n;i++)
        {
            if(num2[i].a==5)
            {
                num2[i].c=100;
            }
            
            else if(num2[i].a==4)
            {
                x++;//记录4出现的次数 
                if(s[4]==1)
                    num2[i].c=95;
                else
                {
                    if(s[4]/2>=x)//如果4出现的次数比中值小 则赋值为95，否则 90 
                    {
                        num2[i].c=95;
                    }
                    else
                    {
                        num2[i].c=90;
                    }
                 }
            }
            else if(num2[i].a==3)
            {
                y++;
                if(s[3]==1)
                    num2[i].c=85;
                else
                {
                    if(s[3]/2>=y)
                    {
                        num2[i].c=85;
                    }
                    else 
                    {
                        num2[i].c=80;
                    }
                 }
            }
            else if(num2[i].a==2)
            {
                z++;
                if(s[2]==1)
                    num2[i].c=75;
                else
                {
                    if(s[2]/2>=z)
                    {
                        num2[i].c=75;
                    }
                    else
                    {
                        num2[i].c=70;
                    }
                 }
            }
            
            else if(num2[i].a==1)
            {
                w++;
                if(s[1]==1)
                    num2[i].c=65;
                else
                {
                    if(s[1]/2>=w)
                    {
                        num2[i].c=65;
                    }
                    else 
                    {
                        num2[i].c=60;
                    }
                 }
            }
            else if(num2[i].a==0)
            {
                num2[i].c=50;
            }
        }
        //转换  根据num2 对num1 中的c赋值 
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(num1[i].a==num2[j].a && num1[i].b==num2[j].b)
                {
                    num1[i].c=num2[j].c;
                    break;
                }
            }
        }
        for(int i=0;i<n;i++)
             cout<<num1[i].c<<endl;
        cout<<endl;
    }
    return 0;
}
//自己造的数据 
/*
15
5 06:30:17
4 07:31:27
4 08:12:12
4 05:23:13
4 06:30:17
3 05:23:13
3 05:23:12
3 05:23:11
3 05:23:10
3 05:23:09
2 06:30:17
2 06:30:12
1 05:23:01
0 08:12:12
*/