本文介绍了一个针对计算机科学专业一年级学生的成绩排名系统。该系统通过比较四门课程（C语言编程、数学、英语及平均分）的成绩来确定每个学生的最佳排名，并强调学生的最高排名以鼓励他们。文章提供了两种实现方案，包括数据输入、排序、排名计算和输出结果的过程。
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
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
ps：题目有些地方说的不是很明确的样子，比如N<=2000，但是从提交的情况看，并没有N==0的情况，所以这里不用考虑。对于排名，若果出现相同分数的科目，应该是按照分数相同排名相同的原则，比如说分数分别为{91,91,90,90,90,89}，那么排名应该是{1,1,3,3,3,6}，而不是简单的{1,2,3,4,5,6}。对于学生id，一般情况应该用string或者是char数组保存，但是本题id是一个六位数，所以就用int保存了，也方便了后面的查找。第一个程序是最先想到并实现的，和第二个程序的差别在于查找id时的方法，第二个程序用了二分查找，代价是需要保持id的递增，但是从提交情况看，时间上明显有很大优势。
第一个程序：
#include<stdio.h> #include<string.h> #include<iostream> #include<vector> #include<algorithm>
using namespace std;
#define N 4
struct Node
{
char id[N];
int grades[N];
int rank[N];
};
int nPeople,nCheck; vector<Node> info;
void Input()
{
scanf("%d%d",&nPeople,&nCheck);
while( nPeople-- )
{
Node tempNode;
scanf("%s%d%d%d",tempNode.id,&tempNode.grades[1],&tempNode.grades[2],&tempNode.grades[3]);
tempNode.grades[0]=tempNode.grades[1]+tempNode.grades[2]+tempNode.grades[3];
info.push_back (tempNode);
}
}
bool cmpA(Node a,Node b)
{
return a.grades[0]>b.grades[0]?true:false;
}
bool cmpC(Node a,Node b)
{
return a.grades[1]>b.grades[1]?true:false;
}
bool cmpM(Node a,Node b)
{
return a.grades[2]>b.grades[2]?true:false;
}
bool cmpE(Node a,Node b)
{
return a.grades[3]>b.grades[3]?true:false;
}
void RankValue(int level)
{
int tempRank=1;
info[0].rank[level]=tempRank;
for(int i=1;i<info.size();i++)
{
if( info[i].grades[level]!=info[i-1].grades[level] )
tempRank=i+1;
info[i].rank[level]=tempRank;
}
}
void Sort()
{
sort(info.begin (),info.end (),cmpA);
RankValue(0);
sort(info.begin (),info.end (),cmpC); RankValue(1);
sort(info.begin (),info.end (),cmpM); RankValue(2);
sort(info.begin (),info.end (),cmpE); RankValue(3); }
int findMax(Node a)
{
int max=0;
for(int i=1;i<N;i++)
if( a.rank[i]<a.rank[max] )
max=i;
return max;
}
void Output()
{
char table[N]={'A','C','M','E'};
while( nCheck-- )
{
char inStr[N];
scanf("%s",inStr);
bool flag=true;
for(int i=0;i<info.size ();i++)
{
if( !strcmp(info[i].id,inStr) )
{
flag=false;
int max=findMax(info[i]);
printf("%d %c\n",info[i].rank[max],table[max]);
break;
}
}
if( flag )
printf("N/A\n");
}
}
int main()
{
Input();
Sort();
Output();
return 0;
}
第二个程序：
#include<stdio.h> #include<string.h> #include<iostream> #include<vector> #include<algorithm>
using namespace std;
#define N 4
struct Node
{
int id;
int grades[N];
int rank[N];
};
int nPeople,nCheck; vector<Node> info;
void Input()
{
scanf("%d%d",&nPeople,&nCheck);
while( nPeople-- )
{
Node tempNode;
scanf("%d%d%d%d",&tempNode.id,&tempNode.grades[1],&tempNode.grades[2],&tempNode.grades[3]);
tempNode.grades[0]=tempNode.grades[1]+tempNode.grades[2]+tempNode.grades[3];
info.push_back (tempNode);
}
}
bool cmpA(Node a,Node b)
{
return a.grades[0]>b.grades[0]?true:false;
}
bool cmpC(Node a,Node b)
{
return a.grades[1]>b.grades[1]?true:false;
}
bool cmpM(Node a,Node b)
{
return a.grades[2]>b.grades[2]?true:false;
}
bool cmpE(Node a,Node b)
{
return a.grades[3]>b.grades[3]?true:false;
}
bool cmpID(Node a,Node b)
{
return a.id<b.id?true:false;
}
void RankValue(int level)
{
int tempRank=1;
info[0].rank[level]=tempRank;
for(int i=1;i<info.size();i++)
{
if( info[i].grades[level]!=info[i-1].grades[level] )
tempRank=i+1;
info[i].rank[level]=tempRank;
}
}
void Sort()
{
sort(info.begin (),info.end (),cmpA);
RankValue(0);
sort(info.begin (),info.end (),cmpC); RankValue(1);
sort(info.begin (),info.end (),cmpM); RankValue(2);
sort(info.begin (),info.end (),cmpE); RankValue(3);
sort(info.begin (),info.end (),cmpID); }
int findMax(Node a)
{
int max=0;
for(int i=1;i<N;i++)
if( a.rank[i]<a.rank[max] )
max=i;
return max;
}
int BinarySearch(int target,int low,int high)
{
int mid=low+(high-low)/2;
while( low<=high )
{
if( target==info[mid].id )
return mid;
if( target>info[mid].id )
low=mid+1;
else
high=mid-1;
mid=(high-low)/2+low;
}
return -1;
}
void Output()
{
char table[N]={'A','C','M','E'};
while( nCheck-- )
{
int inId;
scanf("%d",&inId);
int pos=BinarySearch(inId,0,info.size ());
if( -1!=pos )
{
int max=findMax(info[pos]);
printf("%d %c\n",info[pos].rank[max],table[max]);
}
else
printf("N/A\n");
}
}
int main()
{
Input();
Sort();
Output();
return 0;
}
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