比较学生成绩排名的问题

注意输出格式

成绩比较就是遍历比较

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

实现思想

将id和成绩分开存储,分别放在两个vector中对应下标的位置

存放成绩的时候直接将平均成绩算出并存储

之后读入要比较的学生学号,先和id相比较,不存在则输出N/A

之后比较每个学生和其他学生的成绩大小,比其他人成绩低则排名+1

根据最好排名确定最好的课程

按格式输出排名和课程

代码实现

#include <iostream>
#include <vector>

using namespace std;

vector<string> ids;
vector<int> grades[2000];

int main() {
    int n, m;
    string id;
    int grade_c,grade_m,grade_e, sum, average;
    cin >> n >> m;
    for(int i = 0; i < n; i++) {
        sum = 0;
        cin >> id;
        ids.push_back(id);
        cin >> grade_c >> grade_m >> grade_e;
        average = (grade_c + grade_m + grade_e) / 3;
        grades[i].push_back(average);
        grades[i].push_back(grade_c);
        grades[i].push_back(grade_m);
        grades[i].push_back(grade_e);
    }

    for(int k = 0; k < m; k++) {
        cin >> id;
        int index = -1;
        for(vector<string>::iterator it = ids.begin(); it != ids.end(); it++) {
            if(id == *it){
                index = it - ids.begin();
                break;
            }
        }
        if(index != -1) {
                int A_rank = 1, C_rank = 1, M_rank = 1, E_rank = 1, best_rank = 1; char best_course = ' ';
            for(int j = 0; j < n; j++) {
                if(grades[index][0] < grades[j][0]) {
                    A_rank++;
                }
                if(grades[index][1] < grades[j][1]) {
                    C_rank++;
                }
                if(grades[index][2] < grades[j][2]) {
                    M_rank++;
                }
                if(grades[index][3] < grades[j][3]) {
                    E_rank++;
                }
                best_rank = ((A_rank < C_rank ? A_rank : C_rank) < M_rank ? (A_rank < C_rank ? A_rank : C_rank) : M_rank) < E_rank ? ((A_rank < C_rank ? A_rank : C_rank) < M_rank ? (A_rank < C_rank ? A_rank : C_rank) : M_rank) : E_rank;
                if(best_rank == A_rank) best_course = 'A';
                else if(best_rank == C_rank && best_rank != A_rank) best_course = 'C';
                else if(best_rank == M_rank && best_rank != A_rank && best_rank != C_rank) best_course = 'M';
                else if(best_rank == E_rank && best_rank != A_rank && best_rank != C_rank && best_rank != M_rank) best_course = 'E';
            }
            cout << best_rank << " " << best_course << endl;
        } else {
            cout << "N/A" << endl;
        }
    }
    return 0;
}