题目1002：Grading(九度 Online Judge)

题目1002：Grading(九度 Online Judge)

1，真题

题目1002：Grading

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：10721

解决：2735

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( • 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 thanone 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

来源：

2011年浙江大学计算机及软件工程研究生真题

2，分析

题目大致意思可以解释为如下：

3个人给一个打分，分数分别为G1,G2,G3。其中满分是P,打分误差是T。

代码判断顺序流程如下

若|G1-G2|<=T,则分数是G1和G2的平均分数
若|G1-G2|>T,则第三人给分G3
若G3和G1在误差范围内，但和G2不在，则分数是G1和G3的平均分数
若G3和G2在误差范围内，但和G1不在，则分数是G2和G3的平均分数
若G3和G1以及G2都在误差范围内，则分数是G1,G2,G3中的最大值
若G3和G1以及G2都不在误差范围内，则分数是系统默认分数GJ

3，答案

import java.util.Scanner;
import java.text.DecimalFormat;
/**
 * to test problem_1002
 * @author Sunkun
 * Date: 2013.09.25
 * Memory: 17564KB
 * Code Length: 1145B
 * Time Consuming: 100MS
 */
public class problem_1002 {
	public static void main(String[] args) {
		int P,T,G1,G2,G3,GJ;
		double result;
		int max;
		Scanner input = new Scanner(System.in);
		
		// 具体算法，就不详细解释了
		while(input.hasNext()){
			P = input.nextInt();
			T = input.nextInt();
			G1 = input.nextInt();
			G2 = input.nextInt();
			G3 = input.nextInt();
			GJ = input.nextInt();
		    
			if(T<0 || T>= P || G1<0 || G1>P || G2<0 || G2>P
					|| G3<0 || G3>P || GJ<0 || GJ>P){
				continue;
			}
			if(Math.abs(G2-G1) <= T){
				result =(float)(G1 + G2)/2;
			}else if(Math.abs(G3-G1)<=T && Math.abs(G3-G2)>T){
				result = (float)(G1 + G3)/2;
			}else if(Math.abs(G3-G2)<=T && Math.abs(G3-G1)>T){
				result = (float)(G2 + G3)/2;
			}else if(Math.abs(G3-G1)<=T && Math.abs(G3-G2)<=T){
				max = G1;
				if(G2 > max)
					max = G2;
				if(G3 > max)
					max = G3;
				result = max;
			}else{
				result = GJ;
			}
			
			// 格式化输出
			DecimalFormat df = new DecimalFormat("0.0");
			System.out.println(df.format(result));
		}
	}
}

4，备注

上述代码直接复制是AC不了的，把类名problem_1002改为Main就可以直接上传通过了：）