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;
}