HDU 1084 What Is Your Grade?

What Is Your Grade?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10831    Accepted Submission(s): 3370

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

这道题目用了几个数组去存做了4个，3个，2个，1个的人；反正就是照着题目意思一步步的模拟，谨慎就好

#include<iostream>
#include<algorithm>
#include<string>
#include<map>
#include<stdio.h>
using namespace std;
struct stu
{
	int p,time,mark;
}Num[102];
void t_rank(int *a,int l)
{
	int i,j,t;
	for(i=0;i<l;i++)
	{
		for(j=i+1;j<l;j++)
		{
			if(Num[a[j]].time <Num[a[i]].time )
			{
				t=a[i];
				a[i]=a[j];
				a[j]=t;
			}
		}
	}
	for(i=0;i<l;i++)
		Num[a[i]].mark=i+1;
}
int main()
{
	int a4[102],len4,a3[102],len3,a2[102],len2,a1[102],len1;
	int n,p,h,m,s,i;
	while(cin>>n&&n>=0)
	{
		len4=len3=len2=len1=0;
		for(i=0;i<n;i++){
		scanf_s("%d %d:%d:%d",&p,&h,&m,&s);
		Num[i].p=p;
		Num[i].time=h*3600+m*60+s;
	    switch(p)
		{
		case 4: a4[len4++]=i;break;
		case 3: a3[len3++]=i;break;
		case 2: a2[len2++]=i;break;
        case 1: a1[len1++]=i;break;
		default:break;
		}
		}
		t_rank(a4,len4);
		t_rank(a3,len3);
		t_rank(a2,len2);
		t_rank(a1,len1);
		for(i=0;i<n;i++)
		{
			if(Num[i].p==5) cout<<100<<endl;
			else if(Num[i].p==0) cout<<50<<endl;
			else if(Num[i].p==4)
			{
				if(Num[i].mark<=len4/2) cout<<95<<endl;
				else cout<<90<<endl;
			}
			else if(Num[i].p==3)
			{
				if(Num[i].mark<=len3/2) cout<<85<<endl;
				else cout<<80<<endl;
			}
			else if(Num[i].p==2)
			{
				if(Num[i].mark<=len2/2) cout<<75<<endl;
				else cout<<70<<endl;
			}
			else if(Num[i].p==1)
			{
				if(Num[i].mark<=len1/2) cout<<65<<endl;
				else cout<<60<<endl;
			}
		}
		cout<<endl;
	}
	return 0;
}