九度oj-1002-Grading

2024-06-10 14:18

文章标签 1002 oj 九度 grading

本文主要是介绍九度oj-1002-Grading，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：18522

解决：4789

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

#include<iostream>
#include<iomanip>
#include<algorithm>
using namespace std;
int main()
{int p,t,g1,g2,g3,gj;double g;while(cin>>p>>t>>g1>>g2>>g3>>gj){if(abs(g1-g2)<=t) g=(g1+g2)*1.0/2;else{if(abs(g1-g3)<=t && abs(g2-g3)<=t)g=max(g1,max(g2,g3));else if(abs(g1-g3)<=t || abs(g2-g3)<=t){if(abs(g1-g3)<abs(g2-g3))g=(g1+g3)*1.0/2;else g=(g2+g3)*1.0/2;}else g=gj;}cout<<fixed<<setprecision(1)<<g<<endl;}return 0;
}

这篇关于九度oj-1002-Grading的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！