九度 OJ 1002：Grading

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th是个小屁孩

最新推荐文章于 2021-07-21 22:13:44 发布

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分类专栏：九度OJ 文章标签：九度

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本文介绍了一个用于自动评估大量研究生入学考试试卷的算法。该算法通过三位独立专家对每道题目进行评分，并根据预设的容忍度来确定最终成绩。文章详细解释了不同情况下的评分规则，并提供了一段C++实现代码。

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题目1002：Grading

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：20236

解决：5214

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

来源：

2011年浙江大学计算机及软件工程研究生机试真题

水题，看懂题就行。

#include <iostream>
#include <cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int abs(int n)
{
    return n < 0 ? -n : n;
}
int max(int a,int b)
{
    return a > b ? a : b;
}
int main()
{
    int p, t, g1, g2, g3, gj;
    double grade;
    while(scanf("%d%d%d%d%d%d",&p,&t,&g1,&g2,&g3,&gj)!=EOF)
    {
        if(abs(g1 - g2) <= t)
        {
            grade = (g1 + g2) / 2.0;
        }
        else if(abs(g3 - g2) <= t && abs(g3 - g1) > t)
        {
            grade = (g3 + g2) / 2.0;
        }
        else if(abs(g3 - g1) <= t && abs(g3 - g2) > t)
        {
            grade = (g3 + g1) / 2.0;
        }
        else if(abs(g3 - g1) <= t && abs(g3 - g2) <= t)
        {
            grade = max(g3, max(g1, g2));
        }
        else
        {
            grade = gj;
        }
        printf("%.1lf\n",grade);
    }
}
/**************************************************************
    Problem: 1002
    User: th是个小屁孩
    Language: C++
    Result: Accepted
    Time:0 ms
    Memory:1520 kb
****************************************************************/