A-1012 The Best Rank (25 分)-优快云博客

本文链接：https://blog.youkuaiyun.com/GCWsuger/article/details/113737622

该系统用于评估一年级计算机科学专业学生的成绩，主要考虑C语言、数学（微积分或线性代数）和英语三门课程。学生的表现以其最佳排名为依据，输出每个学生在四类排名（三门课程及平均分）中最好的一项及其排名。输入包含学生数量、查询排名的学生数量、各学生的ID及成绩，输出则是每个查询学生最佳排名及其类别。如果学生不在评分列表中，则显示'N/A'。

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A-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 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
注意点：
大数组要在主函数之外申请

#include <algorithm>
#include <iostream>
using namespace std;
struct hello{
    int num;
    int x[4];
};
int now;
int Rank[1000000][4]={0};
bool cmp(hello a,hello b){
    return a.x[now]>b.x[now];
}
int main(){
    int n,m,i,j,id,min;
    char k[4]={'A','C','M','E'};
    cin >> n >> m;
    hello stu[n];
    for(i=0;i<n;i++){
        cin>>stu[i].num>>stu[i].x[1]>>stu[i].x[2]>>stu[i].x[3];
        stu[i].x[0]=(stu[i].x[3]+stu[i].x[1]+stu[i].x[2]);
    }
    for(now=0;now<4;now++){
        sort(stu,stu+n,cmp);
        Rank[stu[0].num][now]=1;
        for(i=1;i<n;i++){
            if(stu[i].x[now]==stu[i-1].x[now]){
                Rank[stu[i].num][now]=Rank[stu[i-1].num][now];
            }else Rank[stu[i].num][now]=i+1;
        }
    }
    for(j=0;j<m;j++){
        cin >> id;
        if(Rank[id][0]==0)cout << "N/A" <<endl;
        else {
            min=0;
            for(i=0;i<4;i++){
                if(Rank[id][i]<Rank[id][min])min=i;
            }
            cout << Rank[id][min] << ' ' << k[min] <<endl;
        }
    }
    return 0;
}