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PAT A1012. The Best Rank (25)

 阿豪boy
发布于 2017/02/25 16:04
字数 728
阅读 12
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https://www.patest.cn/contests/pat-a-practise/1012

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

To evaluate the performance of our first year CS majored students, we consider their grades of three courses only: C - C Programming Language, M - Mathematics (Calculus or Linear Algebra), and E - English. At the mean time, we encourage students by emphasizing on their best ranks -- that is, among the four ranks with respect to the three courses and the average grade, we print the best rank for each student.

For example, The grades of C, M, E and A - Average of 4 students are given as the following:

StudentID  C  M  E  A
310101     98 85 88 90
310102     70 95 88 84
310103     82 87 94 88
310104     91 91 91 91

Then the best ranks for all the students are No.1 since the 1st one has done the best in C Programming Language, while the 2nd one in Mathematics, the 3rd one in English, and the last one in average.

Input

Each input file contains one test case. Each case starts with a line containing 2 numbers N and M (<=2000), which are the total number of students, and the number of students who would check their ranks, respectively. Then N lines follow, each contains a student ID which is a string of 6 digits, followed by the three integer grades (in the range of [0, 100]) of that student in the order of C, M and E. Then there are M lines, each containing a student ID.

Output

For each of the M students, print in one line the best rank for him/her, and the symbol of the corresponding rank, separated by a space.

The priorities of the ranking methods are ordered as A > C > M > E. Hence if there are two or more ways for a student to obtain the same best rank, output the one with the highest priority.

If a student is not on the grading list, simply output "N/A".

Sample Input

5 6
310101 98 85 88
310102 70 95 88
310103 82 87 94
310104 91 91 91
310105 85 90 90
310101
310102
310103
310104
310105
999999

Sample Output

1 C
1 M
1 E
1 A
3 A
N/A

 

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

struct Stu{
	int id;
	int grade[4]; 	//4个成绩,A C M E  
	int rank[4];
	Stu(){}
	Stu(int id,int a,int c,int m,int e):id(id){
		grade[0]=a;
		grade[1]=c;
		grade[2]=m;
		grade[3]=e;		
	}
	
	bool operator==(const Stu&a){
		return id==a.id;
	} 
	friend ostream& operator<<(ostream& os,const Stu& s){
		return os<<s.id<<" "<<s.grade[0]<<" "<<s.grade[1]<<" "<<s.grade[2]<<" "<<s.grade[3]<<" "<<s.rank[0]<<" "<<s.rank[1]<<" "<<s.rank[2]<<" "<<s.rank[3];
	} 
};
char course[]="ACME";
//cmp函数中使用表示当前按now号成绩排序stu数组 
int now;
bool cmp(Stu a,Stu b){
	return a.grade[now]>b.grade[now];
}


vector<Stu> v;
int main(int argc, char *argv[])
{
	int n,m;
	scanf("%d %d",&n,&m); 
	for(int i=0;i<n;i++){
		int id,a,c,m,e;
		scanf("%d %d %d %d",&id,&c,&m,&e);
		a=round( (c+m+e)/3.0)+0.5;
		Stu t(id,a,c,m,e);
		v.push_back(t);
	}	
	
	//按照各个成绩排序 
	for(now=0;now<4;now++){
		sort(v.begin(),v.end(),cmp);
		v[0].rank[now]=1;
 	 	for(int i=1;i<n;i++){
		  	if(v[i].grade[now]==v[i-1].grade[now])
		  		v[i].rank[now]=v[i-1].rank[now];
	  		else
	  			v[i].rank[now]=i+1;
  		}
		
/*		for(int i=0;i<n;i++)
			cout<<v[i]<<endl;  	
*/	}
	
	int query;
	for(int i=0;i<m;i++){
		scanf("%d",&query);
		Stu s;
		s.id=query;
		vector<Stu>::iterator it = find(v.begin(),v.end(),s);
		if(it==v.end())
			printf("N/A\n");
		else{
			int k,max;
			max=0;	//排名最大的下标  越小越靠前 
			for(k=1;k<4;k++)
				if(it->rank[max]>it->rank[k])
			 		max=k;
			printf("%d %c\n",it->rank[max],course[max]);
		} 
	}
	return 0;
}

 

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