题目1002：Grading

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本文介绍了一个用于研究生入学考试评分的系统逻辑，通过三位独立专家打分，并根据分数差异及容忍范围决定最终成绩的方法。详细解释了不同情况下的评分规则。

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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

<span style="font-family:Courier New;">/*------------------------1002----------------------
description: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].
author     :yuanwei
time     :2013-02-12
--------------------------1002----------------------*/
 
#include <stdio.h>
#include <math.h>
 
//取两个数的最大值
int max(int a, int b)
{
    if (a > b)
    {
        return a;
    }
    else
    {
        return b;
    }
}
 
int main()
{
    int P = 0;
    int T = 0;
    int G1 = 0;
    int G2 = 0;
    int G3 = 0;
    int GJ = 0;
    float average = 0.0;
 
    while (EOF != scanf("%d%d%d%d%d%d", &P, &T, &G1, &G2, &G3, &GJ))
    {
        //判断入参的合法性
        if ((T<0) || (P<0) || (P<=T) || (G1<0) || (G2<0) || (GJ<0) ||
            (G1>P) || (G2>P) || (G3>P) || (GJ>P))
        {
            continue;
        }
 
        // if |G1 - G2|<T, this problem's grade will be the average of G1 and G2.
        if (T >= abs(G1-G2))
        {
            average = (G1 + G2) / 2.0;
        }
        //this problem's grade will be the maximum of the three grades.
        else if ((T>=abs(G1-G3)) && (T>=abs(G2-G3)))
        {
            average = max(G3, max(G1, G2));
        }
        //this problem's grade will be the average of G3 and the closest grade.
        else if (T >= abs(G1-G3))
        {
            average = (G1 + G3) / 2.0;
        }
        //this problem's grade will be the average of G3 and the closest grade.
        else if (T >= abs(G2-G3))
        {
            average = (G2 + G3) / 2.0;
        }
        //a judge will give the final grade GJ
        else
        {
            average = GJ;
        }
 
        printf("%.1f\n", average);
    }
}
 
/**************************************************************
    Problem: 1002
    User: ruoshui1314
    Language: C
    Result: Accepted
    Time:0 ms
    Memory:912 kb
****************************************************************/</span>