题目1002：Grading

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

code

#include <bits/stdc++.h>

using namespace std ;

int main (){

int P ,T , G1,G2,G3,GJ;
float grade ;
while (cin>>P>>T>>G1>>G2>>G3>>GJ){
if (abs(G1-G2)<=T){
grade = 1.0*(G1+G2)/2;
}else {
if (abs(G1-G3)<=T && abs(G2-G3)<=T){
grade = max(G1,max(G2,G3));
}else if (abs(G1-G3)<=T){
grade = 1.0*(G1+G3)/2 ;
}else if (abs(G2-G3)<=T){
grade = 1.0*(G2+G3)/2 ;
}else {
grade = GJ;
}

}
printf("%.1f\n",grade);
}
return 0;
}