九度OnlineJudge之1002：Grading问题的解决方法

题目描述&＃xff1a;

    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(     • 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&＃39;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&＃39;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&＃39;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&＃xff1a;

    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&＃xff1a;

    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&＃xff1a;

20 2 15 13 10 18

样例输出&＃xff1a;        14.0

#include
#include
#include
using namespace std;int max(int a,int b,int c)
{
return a>b?(a>c?a:c):b;
}int main()
{int P,T,G1,G2,G3,GJ;
while(cin>>P>>T>>G1>>G2>>G3>>GJ)
{float grade&＃61;0;if(fabs(G1-G2)<&＃61;T) grade &＃61; (G1&＃43;G2)/2.0;elseif(fabs(G1-G3)<&＃61;T && fabs(G2-G3)>T) grade&＃61; (G3&＃43;G1)/2.0;elseif(fabs(G2-G3)<&＃61;T && fabs(G1-G3)>T) grade&＃61; (G3&＃43;G2)/2.0;elseif(fabs(G1-G3)<&＃61;T && fabs(G2-G3)<&＃61;T) grade &＃61; max(G1,G2,G3);else grade&＃61;GJ;cout<}