JOBDU Q1002

题目：按规则打分。

/*
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:
• 1）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.
• 2）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.
• 3）If G3 is within the tolerance with both G1 and G2, then this problem's grade will be the maximum of the three grades.
• 4）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 <iomanip>	//控制小数点位数时用到的头文件
#include <stdlib.h>
using namespace std;


//判断 if |G1 - G2| ≤ T
bool within_tolerance(int g1, int g2, int t)
{
	int g1_2;
	g1_2 = abs(g1 - g2);
	if (g1_2 < t || g1_2 == t)
		return true;
	else return false;
}

//求两个数的平均数
double avg2(int a, int b)
{
	//特别注意！！为保证double类型的结果的正确性，除数应为2.0，非2！！
	return ((a + b) / 2.0);
}

//求a，b中与o更相近的数，与o的平均数
double avg3(int a, int b, int o)
{
	int x;
	(abs(a - o) < abs(b - o)) ? x = a : x = b;
	return avg2(x, o);
}

//求3个数的最大值
double max3(int a, int b, int c)
{
	int max;
	if (a > b) max = a;
	else max = b;
	if (max < c) max = c;
	return max;
}

int main()
{
	int P, T, G1, G2, G3, GJ;
	double Grade;

	while (cin >> P >> T >> G1 >> G2 >> G3 >> GJ)
	{
		if (within_tolerance(G1, G2, T)) //1）
			Grade = avg2(G1, G2);	
		else if (within_tolerance(G1, G3, T) && within_tolerance(G3, G2, T))	//3）
			Grade = max3(G1, G2, G3);
		else if (!within_tolerance(G1, G3, T) && !within_tolerance(G3, G2, T))	//4）
			Grade = GJ;
		else Grade = avg3(G1, G2, G3);	//2）

		//控制输出一位小数
		cout << setiosflags(ios::fixed);
		cout << setprecision(1) << Grade << endl;
	}
	//system("pause");
	return 0;
}