- 题目描述：
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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：
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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：
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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：
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20 2 15 13 10 18
- 样例输出：
- 14.0
#include <iostream> #include <cmath> #include <iomanip> 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=0; if(fabs(G1-G2)<=T) grade = (G1+G2)/2.0; else if(fabs(G1-G3)<=T && fabs(G2-G3)>T) grade= (G3+G1)/2.0; else if(fabs(G2-G3)<=T && fabs(G1-G3)>T) grade= (G3+G2)/2.0; else if(fabs(G1-G3)<=T && fabs(G2-G3)<=T) grade = max(G1,G2,G3); else grade=GJ; cout<<fixed<<setprecision(1)<<grade<<endl; } //system("pause"); return 0; }