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.

    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].

    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.

20 2 15 13 10 18

14.0

My answer：

#include <iostream>
#include<stdio.h>
#include <stdlib.h>
#include<iomanip>
using namespace std;
int main()
{
	int P, T, G1, G2, G3, GJ;
	float ans;
	cin >> P >> T >> G1 >> G2 >> G3 >> GJ;
	if (abs(G1 - G2) <= T)
        ans = (float)(G1 + G2) / 2;
    else if (abs(G3 - G1) <= T&&abs(G3 - G2)>T)
        ans = (float)(G3 + G1) / 2;
    else if (abs(G3 - G2) <= T&&abs(G3 - G1)>T)
        ans = (float)(G3 + G2) / 2;
    else if (abs(G3 - G2) <= T&&abs(G3 - G1) <= T)
    {
        if(G1>G3)
            ans=(float)G1;
        else if(G2>G1)
            ans=(float)G2;
    }
    else
        ans = (float)GJ;
    cout.setf(ios::fixed);
    cout <<fixed<<setprecision(1)<< ans;
	return 0;
}