九度——题目1002：Grading

最新推荐文章于 2019-03-16 11:22:50 发布

忆瞻

最新推荐文章于 2019-03-16 11:22:50 发布

阅读量1.2k

点赞数

CC 4.0 BY-SA版权

分类专栏：算法

本文链接：https://blog.youkuaiyun.com/wangchangshuai0010/article/details/8639169

算法专栏收录该内容

20 篇文章

订阅专栏

本文探讨了大规模研究生入学考试评分的复杂性及其公平性保障机制。提出了一种评分流程，通过将每道试题分配给三位独立专家进行评估，确保结果尽可能公正。详细阐述了评分规则，包括不同情况下最终得分的确定方式，如当三位专家意见不一致时的处理方法，以及如何根据评分差异和容忍度来决定最终分数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目描述：

    Grading hundreds of thousands of Graduate Entrance Exams is a hard work. It is even harder to design a process to make the results as fair as possible. One way is to assign each exam problem to 3 independent experts. If they do not agree to each other, a judge is invited to make the final decision. Now you are asked to write a program to help this process.
    For each problem, there is a full-mark P and a tolerance T(<P) given. The grading rules are:
    • A problem will first be assigned to 2 experts, to obtain G1 and G2. If the difference is within the tolerance, that is, if |G1 - G2| ≤ T, this problem's grade will be the average of G1 and G2.
    • If the difference exceeds T, the 3rd expert will give G3.
    • If G3 is within the tolerance with either G1 or G2, but NOT both, then this problem's grade will be the average of G3 and the closest grade.
    • If G3 is within the tolerance with both G1 and G2, then this problem's grade will be the maximum of the three grades.
    • If G3 is within the tolerance with neither G1 nor G2, a judge will give the final grade GJ.

输入：

Each input file may contain more than one test case.
Each case occupies a line containing six positive integers: P, T, G1, G2, G3, and GJ, as described in the problem. It is guaranteed that all the grades are valid, that is, in the interval [0, P].

输出：

For each test case you should output the final grade of the problem in a line. The answer must be accurate to 1 decimal place.

样例输入：

20 2 15 13 10 18

样例输出：

14.0

import java.io.BufferedInputStream;
import java.util.Scanner;
 
public class Main
{
    /**
     * @param args
     */
    public static void main(String[] args)
    {
        Scanner cin = new Scanner(new BufferedInputStream(System.in));
        int P, T, G1, G2, G3, GJ;
        double finalGrade;
        while (cin.hasNext())
        {
            P = cin.nextInt();
            T = cin.nextInt();
            G1 = cin.nextInt();
            G2 = cin.nextInt();
            G3 = cin.nextInt();
            GJ = cin.nextInt();
            if (Math.abs(G1 - G2) <= T)
            {
                finalGrade = (double) (G1 + G2) / 2;
            }
            else if ((Math.abs(G3 - G1) <= T) && (Math.abs(G3 - G2) <= T))
            {
                if (G1 > G2)
                {
                    finalGrade = G1;
                }
                else
                {
                    finalGrade = G2;
                }
                if (finalGrade < G3)
                {
                    finalGrade = G3;
                }
            }
            else if (Math.abs(G3 - G1) <= T)
            {
                finalGrade = (double) (G3 + G1) / 2;
            }
            else if (Math.abs(G3 - G2) <= T)
            {
                finalGrade = (double) (G3 + G2) / 2;
            }
            else
            {
                finalGrade = GJ;
            }
            System.out.printf("%.1f\n", finalGrade);
        }
    }
}
 
/**************************************************************
    Problem: 1002
    User: 忆、瞻
    Language: Java
    Result: Accepted
    Time:180 ms
    Memory:17080 kb
****************************************************************/