The Best Rank(测试点分析)

题目

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

答案

#include
#include
#include
#include
using namespace std;
struct Node{
     
	int number; 
	int c,m,e,a;
	int rank[4];
};
bool cmp_a(struct Node tmp1,struct Node tmp2)
{
     
	return tmp1.a>tmp2.a;
}
bool cmp_e(struct Node tmp1,struct Node tmp2)
{
     
	return tmp1.e>tmp2.e;
}
bool cmp_m(struct Node tmp1,struct Node tmp2)
{
     
	return tmp1.m>tmp2.m;
}
bool cmp_c(struct Node tmp1,struct Node tmp2)
{
     
	return tmp1.c>tmp2.c;
}
int main()
{
     
	int n,m,tmp,cnt;
	cin>>n>>m;
	struct Node node[n];
	for(int i=0;i<n;i++)
	{
     
		cin>>node[i].number>>node[i].c>>node[i].m>>node[i].e;
		node[i].a=(node[i].c+node[i].m+node[i].e)/3;
	}
	
	tmp=1;cnt=1;
	sort(node,node+n,cmp_a);
	for(int i=0;i<n;i++)
	{
     
		node[i].rank[0]=tmp;
		if(i!=n-1&&node[i].a!=node[i+1].a) 
		{
     
			tmp+=cnt;cnt=1;
		}
		else cnt++;
	}
	
	tmp=1;cnt=1;
	sort(node,node+n,cmp_c);
	for(int i=0;i<n;i++)
	{
     
		node[i].rank[1]=tmp;
		if(i!=n-1&&node[i].c!=node[i+1].c) 
		{
     
			tmp+=cnt;cnt=1;
		}
		else cnt++;
	}
	
	tmp=1;cnt=1;
	sort(node,node+n,cmp_m);
	for(int i=0;i<n;i++)
	{
     
		node[i].rank[2]=tmp;
		if(i!=n-1&&node[i].m!=node[i+1].m) 
		{
     
			tmp+=cnt;cnt=1;
		}
		else cnt++;
	}
	
	tmp=1;cnt=1;
	sort(node,node+n,cmp_e);
	for(int i=0;i<n;i++)
	{
     
		node[i].rank[3]=tmp;
		if(i!=n-1&&node[i].e!=node[i+1].e) 
		{
     
			tmp+=cnt;cnt=1;
		}
		else cnt++;
	}
	for(int i=0;i<m;i++)
	{
     
		int temp;
		cin>>temp;
		int flag=0;
		for(int j=0;j<n;j++)
		{
     			
			if(node[j].number==temp)
			{
     			
				
				int pos=min_element(node[j].rank,node[j].rank+4)-node[j].rank;
				cout<<*min_element(node[j].rank,node[j].rank+4)<<" ";
				if(pos==0) cout<<'A';
				else if(pos==1) cout<<'C';
				else if(pos==2) cout<<'M';
				else if(pos==3) cout<<'E';
				cout<<endl;
				flag=1;
				break;
			}
		}
		if(flag==0) cout<<"N/A"<<endl;
	}
}

测试点分析

我来说几个没有通过全部测试点的原因：

1. 每个学生的四个排名存在相同的排名，没有按照A > C > M > E的优先级输出
2. 同一科目不同学生的分数相同，排名并列，同时下一个分数不同的同学的排名不一定是前面同学的排名加一。比如分数序列为60 60 60 70 80 80 90，那么其对应的排名应该是1 1 1 4 5 5 7

代码总结

我这个代码其实算是比较长的，原因是我的代码复用部分比较多，但整合成函数又不方便，就只能先这样了。大家思路总体上都是一致的。