九度OJ 题目1002：Grading

最新推荐文章于 2019-03-16 11:22:50 发布

原创最新推荐文章于 2019-03-16 11:22:50 发布 · 550 阅读

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本文介绍了一种研究生入学考试的自动评分系统设计方案。该系统通过三位独立专家进行评分，并根据预设的容忍误差来确定最终成绩。文章提供了一个C++实现示例，详细解释了评分逻辑及特殊情况处理。

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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

我的答案（没考虑优化）：

1.忘了删掉调试用的代码，WR了

2.abs函数

在C下要加上stdlib.h

在C++下，浮点数用#include<cmath>，其它用#include<cstdlib>

3.这题要注意平均后数不是整数的时候，得处理好

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
    int P,T,G1,G2,G3,GJ,m,n;
    double r;
    while(scanf("%d%d%d%d%d%d",&P,&T,&G1,&G2,&G3,&GJ)!=EOF)
    {
        if(abs(G1-G2)<=T)//在容忍范围内
        {
            r=(G1+G2)/2.0;
            printf("%.1lf\n",r);
        }
        else{
            m=abs(G3-G1);
            n=abs(G3-G2);
            if(m<=T||n<=T)
            {
                if(m<=T&&n<=T)     //都接近用最大的那个
                {
                    printf("%.1f\n",max(G1,max(G2,G3))*1.0);
                }else       //最接近的那个
                {
                    printf("%.1lf\n",(m>n?(G3+G2)/2.0:(G3+G1)/2.0));
                }
            }else
            {
                printf("%.1lf\n",GJ*1.0);
            }
        }
    }
    return 0;
}