HDU 1084 - ACM

题目不难，但是需要对数据进行处理，我的代码有些冗长，希望以后能改进。。。

主要思路是先算总的时间，然后进行对比，将做同样题数的前一半的人筛选出来。

/状态：AC/

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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

int M(int i,char T[1005][10])
{
	int n;
	n=(T[i][2]-'0')*36000+(T[i][3]-'0')*3600+(T[i][5]-'0')*600+(T[i][6]-'0')*60+(T[i][8]-'0')*10+(T[i][9]-'0');
	return n;
}

int main()
{
	//freopen("1.txt","r",stdin);
	//freopen("2.txt","w",stdout);
	char T[1005][10]={0};
	long long score[6][1005]={0};
	int c;
	while(cin>>c&&c!=-1)
	{
		getchar();
		int j=0;
		int a0=0,a1=0,a2=0,a3=0,a4=0;
		for(int i=0;i<c;i++){
			for(int m=0;m<10;m++)
				cin.get(T[i][m]);
			getchar();
		}
		for(int i=0;i<c;i++)
		{
			if(T[i][0]=='4')
			{
				score[4][a4++]=M(i,T);
			}
			if(T[i][0]=='3')
			{
				score[3][a3++]=M(i,T);
			}
			if(T[i][0]=='2')
			{
				score[2][a2++]=M(i,T);
			}
			if(T[i][0]=='1')
			{
				score[1][a1++]=M(i,T);
			}
		}
		sort(score[1],score[1]+a1);
		sort(score[2],score[2]+a2);
		sort(score[3],score[3]+a3);
		sort(score[4],score[4]+a4);
		for(int i=0;i<c;i++)
		{
			if(T[i][0]=='4')
			{
				if(a4==1) {cout<<"95"<<endl; continue;}
				for(int k=0;k<a4;k++)
				{
					if(score[4][k]==M(i,T))
					{
						if(k<a4/2) {cout<<"95"<<endl; break;}
						else {cout<<"90"<<endl; break;}
					}
				}
			}
			if(T[i][0]=='3')
			{
				if(a3==1) {cout<<"85"<<endl; continue;}
				for(int k=0;k<a3;k++)
				{
					if(score[3][k]==M(i,T))
					{
						if(k<a3/2) {cout<<"85"<<endl; break;}
						else {cout<<"80"<<endl; break;}
					}
				}
			}
			if(T[i][0]=='2')
			{
				if(a2==1) {cout<<"75"<<endl; continue;}
				for(int k=0;k<a2;k++)
				{
					if(score[2][k]==M(i,T))
					{
						if(k<a2/2) {cout<<"75"<<endl; break;}
						else {cout<<"70"<<endl; break;}
					}
				}
			}
			if(T[i][0]=='1')
			{
				if(a1==1) {cout<<"65"<<endl; continue;}
				for(int k=0;k<a1;k++)
				{
					if(score[1][k]==M(i,T))
					{
						if(k<a1/2) {cout<<"65"<<endl; break;}
						else {cout<<"60"<<endl; break;}
					}
				}
			}
			if(T[i][0]=='5')
			{
				cout<<"100"<<endl; continue;
			}
			if(T[i][0]=='0')
			{
				cout<<"50"<<endl; continue;
			}
		}
		cout<<endl;
	}
	return 0;
}