1012 The Best Rank (25 分)

Posted 57one

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了1012 The Best Rank (25 分)相关的知识,希望对你有一定的参考价值。

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 (≤), 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 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
题目分析:对数据进行处理就可 看了一下柳神的博客 更新相同排名不需要我这么麻烦
比如分数为 80 82 83 83 85
那么需要的排名应该为 1 2 3 3 5
经过处理得到未处理的排名为 1 2 3 4 5
if((*it).Score[i]==(*(it-1)).Score[i])
  (*it).Rank[i]=(*(it-1)).Rank[i];
这样可以了
技术图片
 1 #include<iostream>
 2 #include<vector>
 3 #include<algorithm>
 4 using namespace std;
 5 int flag = 0;
 6 char Project[4] = { C,M,E,A };
 7 typedef struct Score
 8 {
 9     int Student_Id;
10     int score[4];
11     int Rank[4];
12     int High;
13     int Pro;
14 }Scores;
15 bool compare(const Scores& a, const Scores& b)
16 {
17     return a.score[flag] > b.score[flag];
18 }
19 int main()
20 {
21     vector<Scores> S;
22     int N, M;
23     cin >> N >> M;
24     Scores t;
25     for (int i = 0; i < N; i++)
26     {
27         cin >> t.Student_Id >> t.score[0] >> t.score[1] >> t.score[2];
28         t.score[3] = (t.score[0] + t.score[1] + t.score[2]) / 3;
29         S.push_back(t);
30     }
31     for (int i = 0; i < 4; i++)
32     {
33         sort(S.begin(), S.end(), compare);
34         (*S.begin()).Rank[flag] = 0;
35         int j = 0;
36         int truej = 1;
37         for (vector<Scores>::iterator it = S.begin()+1; it != S.end(); it++)
38         {
39             if ((*(it - 1)).score[flag] == (*it).score[flag])
40             {
41                 (*it).Rank[flag] = j;
42                 truej++;
43             }
44             else
45             {
46                 (*it).Rank[flag] = truej;
47                 j = truej++;
48             }
49         }
50         flag++;
51     }
52     for (vector<Scores>::iterator it = S.begin(); it != S.end(); it++)
53     {
54         int min = 65535;
55         int minp = 0;
56         for (int i = 0; i < 3; i++)
57         {
58             if (min > (*it).Rank[i])
59             {
60                 min = (*it).Rank[i];
61                 minp = i;
62             }
63         }
64         if ((*it).Rank[3] <=min)
65         {
66             min = (*it).Rank[3];
67             minp = 3;
68         }
69         (*it).High = min+1;
70         (*it).Pro = minp;
71     }
72     for (int i = 0; i < M; i++)
73     {
74         cin >> t.Student_Id;
75         int flag =0;
76         for (vector<Scores>::iterator it = S.begin(); it != S.end(); it++)
77             if ((*it).Student_Id == t.Student_Id)
78             {
79                 cout << (*it).High << " " << Project[(*it).Pro] << endl;
80                 flag = 1;
81                 break;
82             }
83         if (!flag)
84             cout << "N/A"<<endl;
85     }
86     return 0;
87 }
View Code

 

以上是关于1012 The Best Rank (25 分)的主要内容,如果未能解决你的问题,请参考以下文章

1012 The Best Rank (25 分)

1012 The Best Rank (25分) vector与结构体排序

PAT Advanced 1012 The Best Rank (25分)

[PTA] PAT(A) 1012 The Best Rank (25 分)

PAT-1012 The Best Rank (25 分) 查询分数对应排名(包括并列)

1012 The Best Rank (25 分)难度: 中 / 知识点: 排序 前缀和