PAT甲级 1012 The Best Rank (25)

1012 The Best Rank (25 分)

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 CME 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 Nlines 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 CM 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

 思路,首先要想到他会查询没有的id,一开始我想到的是用vector,但是内存超了,后面改用int二维数组,然后鉴于每次都要排序计算排名,一共排4次,那么可以把成绩保存到一个int数组中,根据下标的选择写一个for循环来控制每次应该排哪个,这样减少了重复的代码量,最后把排名保存到int二维数组中,每次查询id时遍历二维数组找到最高的排名,根据题意还有排名相同根据题目自定义排序输出,我们初始下标就默认0为A,1为C,2为M,3为E,那么判断起来就方便了许多

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;

const int inf=1000000000;
struct peo{
	int name;
	int grade[4];
}P[2010];

char arr[4]={'A','C','M','E'};
int v[1000000][4];
int now;

int roundd(int x){
	double aver=x/3.0;
	int a=x/3;
	if(aver-a>=0.5) return a+1;
	else return a;
}
bool cmp(peo p1,peo p2){
	return p1.grade[now]>p2.grade[now];
}
int main(){
	int n,k,id;
	scanf("%d %d",&n,&k);
	peo p;
	for(int i=0;i>id;
		if(v[id][0]==0) printf("N/A\n");
		else{
			int minrank=inf,minindex=inf;
			for(int j=0;j<4;j++){
				if(v[id][j]

 

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