题目1002：Grading

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：22664

解决：5821

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:     • 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：

    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> #include<cmath> #include<algorithm> using namespace std; int main(){ double P,T,G1,G2,G3,GJ; while(cin>>P>>T>>G1>>G2>>G3>>GJ){ if(G1>P||G1<0||G2>P||G2<0||G3>P||G3<0||GJ>P||GJ<0) return false; if(abs(G1-G2)<=T){ cout<<fixed<<setprecision(1)<<(G1+G2)/2.0<<endl; } else if((abs(G3-G1)<=T)&&(abs(G3-G2)>T)){ cout<<fixed<<setprecision(1)<<(G1+G3)/2.0<<endl; } else if((abs(G3-G2)<=T)&&(abs(G3-G1)>T)){ cout<<fixed<<setprecision(1)<<(G2+G3)/2.0<<endl; } else if(abs(G3-G1)>T&&abs(G3-G2)>T){ cout<<fixed<<setprecision(1)<<GJ<<endl; } else{ double a[3]={G1,G2,G3}; sort(a,a+3); cout<<a[2]<<endl; } } return 0; }