PAT 甲级 1012 The Best Rank

1012 The Best Rank （25 point(s)）

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 Algrbra), 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 Specification:

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 Specification:

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

Experiential Summing-up

This question can not be solved by sorting structure array. You would get a result of time limit exceeded if you submit code with structure array. The mainly reason is student ID which is a string of six digits. That determine the size of array must greater than 10^6, while the time limit is 200ms and computer's times of calculation is 10^7~10^8/s, which means two times of sort will cost your all time allowed. So don't imagine use array of structure to cope with it. STL vector is a good selection~As for methods of sort. It's easy to complete it. You can!~

(The purpose of using English to portray my solution is that to exercise the ability of my expression of English and accommodate PAT advanced level's style.We can make progress together by reading and comprehending it. Please forgive my basic grammar's and word's error. Of course, I would appreciated it if you can point out my grammar's and word's error in comment section.( •̀∀•́ ) Furthermore, Big Lao please don't laugh at me because I just a English beginner settle for CET6    _(:з」∠)_  )

Accepted Code

#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
const int INF=0x3fffffff;
int n,f;
struct student
{
	int rank[4];
	int no;
	int flag;
	int score[4];
};
vector<student> stu;
bool cmp1(student a,student b)
{
	return a.score[f]>b.score[f];
}
bool cmp2(student a,student b)
{
	return a.no<b.no;
}
void ranking()
{
	stu[0].rank[f]=1;
	for(int i=1;i<n;++i)
	{
		stu[i].rank[f]=i+1;
		if(stu[i].score[f]==stu[i-1].score[f])
			stu[i].rank[f]=stu[i-1].rank[f];
	}
}
void judge(int id,int &rank,int &index)
{
	rank=INF;
	index=0;
	for(int i=0;i<4;++i)
	{
		if(stu[id].rank[i]<rank)
		{
			rank=stu[id].rank[i];
			index=i;
		}
	}
}
int search(int no)
{
	int l=0,h=n-1;
	while(l<=h)
	{
		int mid=(l+h)/2;
		if(stu[mid].no==no)
			return mid;
		else if(stu[mid].no<no)
			l=mid+1;
		else 
			h=mid-1;
	}
	return -1;
}
int main()
{
	int m,no;
	char s[4]={'A','C','M','E'};
	scanf("%d %d",&n,&m);

	for(int i=0;i<n;++i)
	{
		student a;
		scanf("%d",&no);
		a.flag=1;
		a.no=no;
		scanf("%d %d %d",&a.score[1],&a.score[2],&a.score[3]);
		a.score[0]=(a.score[1]+a.score[2]+a.score[3])/3;
		stu.push_back(a);
	}
	for(f=0;f<4;++f)
	{
		sort(stu.begin(),stu.end(),cmp1);
		ranking();
	}
	sort(stu.begin(),stu.end(),cmp2);
	for(int i=0;i<m;++i)
	{
		scanf("%d",&no);
		int rank=INF,index=0;
		int x=search(no);
		if(x==-1)
			printf("N/A\n");
		else
		{
			judge(x,rank,index);
			printf("%d %c\n",rank,s[index]);
		}
	}
	return 0;
}