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

#include <iostream>  #include <algorithm>#include <cstdlib>   //abs函数using namespace std;  int main(){int P,T,G1,G2,G3,GJ;while(cin >> P >> T >> G1 >> G2 >> G3 >> GJ){double finalscore;if(abs(G1 - G2) <= T)finalscore = (G1 + G2) / 2.0;else if(abs(G1 - G3) <= T && abs(G2 - G3) <= T)finalscore = max(max(G1,G2),G3);else if(abs(G1 - G3) > T && abs(G2 - G3) > T)finalscore = GJ;else{if(abs(G1 - G3) > abs(G2 - G3))   // 这里的意思是说abs(G1 - G3) > T && abs(G2 - G3) <= Tfinalscore = (G2 + G3) / 2.0;else  // 这里的意思是说abs(G2 - G3) > T && abs(G1 - G3) <= Tfinalscore = (G1 + G3) / 2.0;}cout.setf(ios::fixed);cout.precision(1);cout << finalscore << endl;}return 0;}