What Is Your Grade?(水，排序)

 

What Is Your Grade?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9935    Accepted Submission(s): 3041

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

题解：虽然水题，但是还是贴一下吧，毕竟刚开始没读清题错了几次，贴下长个记性；刚开始读成相同题目的第一名+5了，最后又读了下，发现是前一半的+5。。。

对了，还有像time，rank了都不能在代码里面命名否则ce；

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",&x)
#define P_ printf(" ")
int ans[1010];
int nu[6];
struct Node{
	int cnt;
	int num;
	int hh,mm,dd;
	int sec;
	Node init(int x){
		cnt=x;
		scanf("%d",&num);
		nu[num]++;
		scanf("%d:%d:%d",&hh,&mm,&dd);
		sec=hh*60*60+mm*60+dd;
		ans[cnt]=50+10*num;
	}
	friend bool operator < (Node a,Node b){
		if(a.num!=b.num)return a.num<b.num;
		else return a.sec<b.sec;
	}
}dt[1010];
int main(){
	int N;
	while(SI(N),N!=-1){
		mem(nu,0);
		for(int i=0;i<N;i++){
			dt[i].init(i);
		}
		sort(dt,dt+N);
		for(int i=1;i<=4;i++){
			for(int j=0;j<N;j++){
				if(dt[j].num==i){
					if(nu[i]==1)ans[dt[j].cnt]+=5;
					else for(int k=j;k<j+nu[i]/2;k++){
						ans[dt[k].cnt]+=5;
					}
					break;
				}
			}
		}
		for(int i=0;i<N;i++){
			printf("%d\n",ans[i]);
		}
		puts("");
	}
	return 0;
}