1002.Grading

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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

#include<iostream>
#include<iomanip> //cout<<fixed<<setprecision(1)<<  !!!! 
using namespace std;

int cha(int x,int y){
    if(x>y) return x-y;
    else return y-x;
}

int maximum(int a,int b,int c)
{
    if(a>b){
        if(a>c) return a;
        else return c;
    }
    else {
        if(b>c) return b;
        else return c;
    }
}


int main(){
    int p,t,g1,g2,g3,gj;
    double score;
    while(cin>>p>>t>>g1>>g2>>g3>>gj){
        if(cha(g1,g2)<=t) {
            score=(g1+g2)/2;
        }
        else if(cha(g1,g3)>t && cha(g2,g3)>t){
            score=gj;
        }
        else if(cha(g1,g3)<=t && cha(g2,g3)<=t){
            score=maximum(g1,g2,g3);
        }
        else if(cha(g1,g3)<cha(g2,g3)){
            score=(g1+g3)/2;
        }
        else score=(g2+g3)/2;
        cout<<fixed<<setprecision(1)<<score<<endl;
    }
    return 0;
}

 

1002.Grading

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原文地址：https://www.cnblogs.com/bernieloveslife/p/9736598.html