九度题目1324：The Best Rank - 程序猿之洞 - 博客频道 - CSDN.NET
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.
输入：
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.
输出：
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".
样例输入：
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样例输出：
1 C
1 M
1 E
1 A
3 A
N/A
Read full article from 九度题目1324：The Best Rank - 程序猿之洞 - 博客频道 - CSDN.NET
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.
输入：
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.
输出：
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".
样例输入：
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样例输出：
1 C
1 M
1 E
1 A
3 A
N/A
- struct node
- {
- int ID,C,M,E,A;//学生的编号，三科的成绩和平均成绩
- int PC,PM,PE,PA;//这几科的排名
- int con;//最高排名的成绩
- char val;//最高排名的课的名字
- }stu[2500];
06 | struct stu { |
07 | stu(string& n,int i,int j,int k):id(n) { |
08 | c[0]=(i+j+k)/3; |
09 | c[1]=i;c[2]=j;c[3]=k; |
10 | } |
11 | string id; |
12 | int c[4],bst,tp,r[4]; |
13 | }; |
14 | bool p1(const stu a,const stu b) { |
15 | if(a.c[0]>b.c[0])return true; |
16 | else return false; |
17 | } |
18 | bool p2(const stu a,const stu b) { |
19 | if(a.c[1]>b.c[1])return true; |
20 | else return false; |
21 | } |
22 | bool p3(const stu a,const stu b) { |
23 | if(a.c[2]>b.c[2])return true; |
24 | else return false; |
25 | } |
26 | bool p4(const stu a,const stu b) { |
27 | if(a.c[3]>b.c[3])return true; |
28 | else return false; |
29 | } |
30 | int main() { |
31 | int n,m,c,e,im; |
32 | string id; |
33 | bool (*cp[4])(const stu,const stu); |
34 | cp[0]=p1;cp[1]=p2;cp[2]=p3;cp[3]=p4; |
35 | while(cin>>n>>im) { |
36 | vector < stu > vs; |
37 | while(n--) { |
38 | cin>>id>>c>>m>>e; |
39 | vs.push_back(stu(id,c,m,e)); |
40 | } |
41 | for(int fc=0; fc<4; ++fc) { |
42 |
43 | sort(vs.begin(),vs.end(),cp[fc]); |
44 | vs[0].r[fc]=1; |
45 | for(int i=1; i<vs.size(); ++i) |
46 | if(vs[i].c[fc]==vs[i-1].c[fc]) |
47 | vs[i].r[fc]=vs[i-1].r[fc]; |
48 | else vs[i].r[fc]=i+1; |
49 | } |
50 | for(int i=0;i<vs.size();++i){ |
51 | vs[i].bst=vs[i].r[0];vs[i].tp=0; |
52 | for(int k=1;k<4;++k) |
53 | if(vs[i].r[k]<vs[i].bst){ |
54 | vs[i].bst=vs[i].r[k]; |
55 | vs[i].tp=k; |
56 | } |
57 | } |
58 | while(im--) { |
59 | bool ok=false; |
60 | cin>>id; |
61 | for(int i=0; i<vs.size(); ++i) { |
62 | if(vs[i].id==id) { |
63 | ok=true; |
64 | cout<<vs[i].bst<<' '; |
65 | switch(vs[i].tp) { |
66 | case 0:cout<<"A"<<endl;break; |
67 | case 1:cout<<"C"<<endl;break; |
68 | case 2:cout<<"M"<<endl;break; |
69 | case 3:cout<<"E"<<endl;break; |
70 | } |
71 | break; |
72 | } |
73 | } |
74 | if(!ok)cout<<"N/A"<<endl; |
75 | } |
76 | } |
77 | } |
