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Jobdu 1002:Grading  

2012-08-30 20:36:03|  分类: ACM |  标签: |举报 |字号 订阅

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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
/////////////以下是个人的粗糙代码,有待改进,仅供参考
 
#include <iostream>
#include <cstdlib>
#include <iomanip>
using namespace std;
int main()
{
   int P,T,G1,G2,G3,GJ;
   double result;
   while((cin>>P>>T>>G1>>G2>>G3>>GJ))
   {
     if(abs(G1-G2)<=T)
     {
         result=((double)(G1+G2))/2;
         cout<<setiosflags(ios::fixed);
         cout<<setprecision(1)<<result<<endl;
     }
     else
     {
        if(abs(G1-G3)<=T&&abs(G2-G3)>T)
        {
           result=((double)(G1+G3))/2;
           cout<<setiosflags(ios::fixed);
           cout<<setprecision(1)<<result<<endl;
        }
        else if(abs(G1-G3)>T&&abs(G2-G3)<=T)
        {
           result=((double)(G2+G3))/2;
           cout<<setiosflags(ios::fixed);
           cout<<setprecision(1)<<result<<endl;
        }
        else if(abs(G1-G3)<=T&&abs(G2-G3)<=T)
        {
           result=(double)(G1>(G2>G3? G2:G3)? G1:(G2>G3? G2:G3));
           cout<<setiosflags(ios::fixed);
           cout<<setprecision(1)<<result<<endl;
        }
        else
        {
           cout<<setiosflags(ios::fixed);
           cout<<setprecision(1)<<(double)GJ<<endl;
        }
     }
   }
   return 0;
}
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