_jobdu_1002

 /************************************************************************/
 /* 题目1002：Grading
     时间限制：1 秒
     内存限制：32 兆
     特殊判题：否

     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:
     P满分、T公差。
     • 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.
     ·如果G1和G2的差小于公差，则取平均。
     • If the difference exceeds T, the 3rd expert will give G3.
     ·如果G1和G2差大于公差，则看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.
     ·如果G3与G1或G2中一个差小于公差，则取G3及相近一成绩的平均。
     • If G3 is within the tolerance with both G1 and G2,
     then this problem's grade will be the maximum of the three grades.
     ·如果G3与G1、G2差均小于公差。则最终成绩取最大值。
     • If G3 is within the tolerance with neither G1 nor G2,
     a judge will give the final grade GJ.
     ·如果G3与G1、G2差值均大于公差，则由GJ决定最终成绩。
     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>
 using namespace std;
 double sAbs(double x)
 {
     if(x>=)return x;
     else return -x;
 }
 double max(double x, double y , double z)
 {
     return
     (x>y)?
         ((x>z)?x:z)
         :((y>z)?y:z);
 }
 int main()
 {
     double P,T,G1,G2,G3,GJ;
     double temp,temp3,final;
     bool   isTwoLarger;
     P = T = G1 = G2 = G3 = GJ = ;
     isTwoLarger = false;
     while(cin>>P>>T>>G1>>G2>>G3>>GJ)
     {
         if(T>P)cout<<"Data Error!"<<endl;
         //((G1-G2)<=0)?(isTwoLarger = true):(isTwoLarger = false);
         //if(isTwoLarger)temp = G2 - G1;
         //else temp = G1 - G2;
         //if(temp<=T)final = (G1+G2)/2.0;
         //else if(isTwoLarger)
         //{
         //    if(G3<=G1)
         //        if((G1-G3)>T)final = GJ;
         //        else final = (G3+G1)/2.0;
         //    else if(G3>G2)
         //        if((G3-G2)>T)final = GJ;
         //        else final = (G3+G2)/2.0;
         //    else
         //    {
         //        temp = G3-G1;
         //        temp3 = G2-G3;
         //        if((temp<=T)&&(temp3<=T))final = G2;
         //        else if((temp>T)&&(temp3>T))final = GJ;
         //        else if((temp3-temp)>0)
         //            final = (G3+G1)/2.0;
         //        else final = (G3+G2)/2.0;
         //    }
         //}
         //else
         //{
         //    if(G3<=G2)
         //        if((G2-G3)>T)final = GJ;
         //        else final = (G3+G2)/2.0;
         //    else if(G3>G1)
         //        if((G3-G1)>T)final = GJ;
         //        else final = (G3+G1)/2.0;
         //    else
         //    {
         //        temp = G3-G2;
         //        temp3 = G1-G3;
         //        if((temp<=T)&&(temp3<=T))final = G1;
         //        else if((temp>T)&&(temp3>T))final = GJ;
         //        else if((temp3-temp)>0)
         //            final = (G3+G2)/2.0;
         //        else final = (G3+G1)/2.0;
         //    }
         //}
         if(sAbs(G1-G2)<=T)final = (G1+G2)/2.0;
         else if((sAbs(G1-G3)<=T)&&(sAbs(G2-G3)<=T)) final = max(G1,G2,G3);
         else if(sAbs(G2-G3)<=T) final = (G2+G3)/2.0;
         else if(sAbs(G1-G3)<=T)final = (G1+G3)/2.0;
         else final = GJ;
         cout<<fixed<<::setprecision()<<final<<endl;
         //printf("%.1f\n",final);
     }
     return ;
 }

 //#include <iostream>
 //#include <iomanip>
 //#include <cmath>
 //using namespace std;
 //double getMax( double x, double y, double z)
 //{
 //    return
 //    (x>y)?
 //        ((x>z)?x:z)
 //        :((y>z)?y:z);
 //
 //}
 //int main()
 //{
 //    double P,T,G1,G2,G3,GJ;
 //    double final;
 //    cin>>P>>T>>G1>>G2>>G3>>GJ;
 //    if(abs(G1-G2)<=T)
 //    {
 //        cout<<fixed<<::setprecision(1)<<(G1+G2)/2.0<<endl;
 //    }
 //    else if(abs(G3-G1)<=T&&abs(G3-G2)<=T)
 //    {
 //        cout<<fixed<<::setprecision(1)<<getMax(G1,G2,G3)<<endl;
 //    }
 //    else if(abs(G3-G2)<=T)
 //    {
 //        cout<<fixed<<::setprecision(1)<<(G3+G2)/2.0<<endl;
 //    }
 //    else if(abs(G3-G1)<=T)
 //    {
 //        cout<<fixed<<::setprecision(1)<<(G3+G1)/2.0<<endl;
 //    }
 //    else
 //    {
 //        cout<<fixed<<::setprecision(1)<<GJ<<endl;
 //    }
 //    return 0;
 //
 //}