1012 The Best Rank (25 分)

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

Output Specification:

For each of the

The priorities of the ranking methods are ordered as A

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

Submit:

#include <iostream>
#include <algorithm>

using namespace std;
//目标：对四个分数从高到底排序，每个考生4个排名，输出当前学生最好的排名和对应分数，名次相同 A>C>M>E
struct node{
    int id,best;
    int score[4],rank[4];
}stu[2005];
int exist[1000000], flag = -1;
bool cmpl(node a, node b) {return a.score[flag] > b.score[flag];}
int main(){
    int n,m,id;
    scanf("%d %d",&n,&m);
    for (int i=0;i<n;i++) {
        scanf("%d %d %d %d",&stu[i].id,&stu[i].score[1],&stu[i].score[2],&stu[i].score[3]);
        stu[i].score[0] = (stu[i].score[1]+stu[i].score[2]+stu[i].score[3])/3.0 + 0.5;//+0.5取证 保证四舍五入
    }
    for (flag = 0;flag <=3; flag++) {//排序
        sort(stu,stu+n,cmpl);
        stu[0].rank[flag] = 1;
        for (int i=1;i<n;i++) {
            stu[i].rank[flag] = i+1;
            if (stu[i].score[flag]==stu[i-1].score[flag])
                stu[i].rank[flag] = stu[i-1].rank[flag];
        }
    }
    for (int i=0;i<n;i++) {
        exist[stu[i].id] = i+1;//把有成绩的学生id存入exist数组中，m行id没有存进来表示为0输出N/A
        stu[i].best = 0;
        int minn = stu[i].rank[0];
        for (int j=1;j<=3;j++) {//找出成绩最小排名即最好成绩
            if (stu[i].rank[j] < minn) {
                minn = stu[i].rank[j];
                stu[i].best = j;//位置
            }
        }
    }
    char c[5] = {'A','C','M','E'};
    for (int i=0;i<m;i++) {
        scanf("%d",&id);
        int temp = exist[id];
        if(temp){
            int best = stu[temp-1].best;//temp-1是该学生id对应的下标
            printf("%d %c
",stu[temp-1].rank[best], c[best]);
        } else{
            printf("N/A
");
        }
    }
    return 0;
}

参考：

柳婼-https://blog.csdn.net/liuchuo/article/details/54561626