• 题目1002:Grading


    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

    Java Code
    import java.util.Scanner;
    
    public class Main{   
        public static void main(String[] args) {
            Scanner  in = new  Scanner(System.in);
            while(in.hasNextInt()){
                int P = in.nextInt();
                int T = in.nextInt();
                int G1 = in.nextInt();
                int G2 = in.nextInt();
                int G3 = in.nextInt();
                int GJ = in.nextInt();
                
                float grade = 0 ;
                
                int a = Math.abs(G1 - G2);
                int b = Math.abs(G1 - G3);
                int c = Math.abs(G2 - G3);
                
                if(a <= T)
                    grade = (float)(G1 + G2)/2;
                else{
                    if((b <= T) && (c <= T)){
                        int temp = G1 > G2 ? G1 : G2;    
                        grade = temp > G3 ? temp :G3;
                    }
                    
                    if((b > T) && (c > T))
                        grade = GJ;
                    if(b <= T)
                        grade = (float)(G1 + G3)/2;
                    if(c <= T)
                        grade = (float)(G2 + G3)/2;                    
                }
                System.out.println(grade);
                
            }
        }
    
    }
  • 相关阅读:
    .NET微服务调查结果
    发布基于Orchard Core的友浩达科技官网
    Followme Devops实践之路
    积极参与开源项目,促进.NET Core生态社区发展
    Service Fabric 与 Ocelot 集成
    “.Net 社区大会”(dotnetConf) 2018 Day 1 主题演讲
    Project file is incomplete. Expected imports are missing 错误解决方案
    稳定工作和创业之间的抉择
    回顾4180天在腾讯使用C#的历程,开启新的征途
    ML-Framework:ML.NET 0.3 带来新组件
  • 原文地址:https://www.cnblogs.com/Mokaffe/p/4065644.html
Copyright © 2020-2023  润新知