PATA1012The Best Rank（25分）

秒速五厘米 2021-11-02 08:34 234阅读 0赞

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

### 开始技术总结 ###

*  这一题是个排序题，分为两块首先是结构体的设计，有id记录学号注意是整型不是字符型和再就是一个grade整型数组，记录平均分和其他三门成绩。
 *  在通过额外定义一个整型Rank数组二维，一维记录学号，二维是4个空间记录每一门成绩的排名。
 *  同时定义了一个全局变量now，用于函数cmp，里面的参数使用。这里有一个坑，就是在下文中可以看见，我是额外的通过在for循环定义了j，然后再是now=j，而不是直接的在for循环里面定义now，这里是我一个错误点，导致纠结了好久。应该是for循环里面的变量是不能充当全局变量使用了。
 *  在就是定义一个number数组来存储，后面需要输出排名学生的学号，不能输入一个就输出一个排名。
 *  接下来就是每一位学生每一科排名的处理，重点因为是4科成绩，所以for循环四次，然后每次通过sort函数排序，排好之后每次给第一位学生的该门成绩排名为1，之后，再是给剩下的学生排名，如果成绩相同，那么排名就与之前的学生排名相同，如果不是就排名为i+1，而不是在原来排名的基础上加1。这部分代码如下：

for (int j = 0; j < 4; j++) {
            now = j;
            sort(stu, stu + n, cmp);
            Rank[stu[0].id][now] = 1;
            for (int i = 1; i < n; i++) {
                if (stu[i].grade[now] == stu[i - 1].grade[now]) {//如果分数相同则排名相同，与前一位进行比较
                    Rank[stu[i].id][now] = Rank[stu[i-1].id][now];
                }
                else {
                    Rank[stu[i].id][now] = i + 1;
                }
            }
    
        }

*  然后就是输出部分了，有了之前排好名的Rank数组，如果Rank数组的二维第一位排名为0，说明该学号学生不存在。直接输出N/A，然后如果不为0，那么就需要在4个中寻找排名最靠前的，如果有多个排名相同，按照A>C>M>E的优先级输出一个即可。这里也是为啥在定义结构体Student和数组Rank时，按照0~3，存放的分数和排名顺序是按照这个优先级来的。方便输出。这部分参考代码如下：

for (int i = 0; i < m; i++) {
            if (Rank[number[i]][0] == 0) {
                printf("N/A\n");
            }
            else {
                int k = 0;//用于选出Rank[number[i]]中最小的
                for (int j = 0; j < 4; j++) {
                    if (Rank[number[i]][j] < Rank[number[i]][k]) {
                        k = j;
                    }
                }
                printf("%d %c\n", Rank[number[i]][k], course[k]);
            }
            
        }

#### 所有部分参考代码入下: ####

//在VS2017上编写，在PAT上通过测试25分满的
    #define _CRT_SECURE_NO_WARNINGS
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    using namespace std;
    
    struct Student {
        int id;
        int grade[4];
    }stu[2010];
    
    char course[4] = { 'A','C','M','E' };
    int Rank[1000000][4];//用于记录每位学生的每门成绩排名,且从0到3分别表示A，C，M，E的成绩排名
    int now;//记录是哪一门成绩开始比较
    
    bool cmp(Student a, Student b) {
        return a.grade[now] > b.grade[now];
    }
    
    int main() {
        int m, n;
        scanf("%d%d", &n, &m);
        for (int i = 0; i < n; i++) {
            scanf("%d%d%d%d", &stu[i].id, &stu[i].grade[1], &stu[i].grade[2], &stu[i].grade[3]);
            stu[i].grade[0] = round((stu[i].grade[1] + stu[i].grade[2] + stu[i].grade[3]) / 3.0);//求出平均分即A
        }
    
        int number[2010];//用来存储需要输出的学号
        for (int i = 0; i < m; i++) {
            scanf("%d", &number[i]);
        }
    
        for (int j = 0; j < 4; j++) {
            now = j;
            sort(stu, stu + n, cmp);
            Rank[stu[0].id][now] = 1;
            for (int i = 1; i < n; i++) {
                if (stu[i].grade[now] == stu[i - 1].grade[now]) {//如果分数相同则排名相同，与前一位进行比较
                    Rank[stu[i].id][now] = Rank[stu[i-1].id][now];
                }
                else {
                    Rank[stu[i].id][now] = i + 1;
                }
            }
    
        }
    
        for (int i = 0; i < m; i++) {
            if (Rank[number[i]][0] == 0) {
                printf("N/A\n");
            }
            else {
                int k = 0;//用于选出Rank[number[i]]中最小的
                for (int j = 0; j < 4; j++) {
                    if (Rank[number[i]][j] < Rank[number[i]][k]) {
                        k = j;
                    }
                }
                printf("%d %c\n", Rank[number[i]][k], course[k]);
            }
            
        }
    
        system("pause");
        return 0;
    }

转载于:https://www.cnblogs.com/tsruixi/p/11327661.html