PAT 1012 The Best Rank_score":85},{"averageline":27,"rank":61,"rankname":-优快云博客

本文介绍了一个学生成绩排名系统的设计与实现。该系统针对计算机科学专业一年级学生的三门课程（C语言编程、数学、英语）成绩进行排名，并强调每位学生的最佳排名。通过样例输入输出展示了系统的运行方式。

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1012. The Best Rank (25)

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

//把每科的成绩情况存到结构体数组，数组空间动态分配
//每组调用sort函数进行排序，提前对结构体重载<运算符
//注意：
//对于rank
//如：94 86 86 85 85 78
//rank应为1 2 2 4 4 6，而不是1 2 2 3 3 4
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
struct node{
	string name;
	int score;
	bool operator <(const node& b){
		return this->score>b.score;
	}
}*sub[4];
int main(){
	int n,m,score,a;
	char re[5]="ACME";
	int top[4]={-1,-1,-1,-1};
	string names;
	cin>>n>>m;
	string *name=new string[n+5];
	for(int i=0;i<4;i++){
		sub[i]=new node[n+5];
	}
	for(int i=1;i<=n;i++){
		cin>>name[i];
		a=0;
		for(int j=1;j<4;j++){
			cin>>score;
			a+=score;
			++top[j];
			sub[j][top[j]].name=name[i];
			sub[j][top[j]].score=score;
		}
		a/=3;
		++top[0];
		sub[0][top[0]].name=name[i];
		sub[0][top[0]].score=a;
	}
	for(int i=0;i<4;i++){
		sort(sub[i],sub[i]+top[i]+1);
	}
	for(int i=1;i<=m;i++){
		cin>>names;
		int flag=0;
		for(int j=1;j<=n;j++){
			if(name[j]==names){
				flag=1;
				break;
			}
		}
		if(!flag){
			cout<<"N/A"<<endl;
			continue;
		}
		int step=n+1,index;
		for(int j=0;j<4;j++){
			int rank=0;
			for(int t=0;t<=top[j];t++){
				if(t==0||sub[j][t].score!=sub[j][t-1].score){
					rank=t+1;
				}
				if(names==sub[j][t].name){
					if(rank<step){
						step=rank;
						index=j;
						break;
					}
				}
			}
		}
		cout<<step<<' '<<re[index]<<endl;
	}
	delete [] name;
	for(int i=0;i<4;i++){
		delete [] sub[i];
	}
	return 0;
}