1002.Grading
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- 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:
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14.0
#include<iostream> #include<iomanip> //cout<<fixed<<setprecision(1)<< !!!! using namespace std; int cha(int x,int y){ if(x>y) return x-y; else return y-x; } int maximum(int a,int b,int c) { if(a>b){ if(a>c) return a; else return c; } else { if(b>c) return b; else return c; } } int main(){ int p,t,g1,g2,g3,gj; double score; while(cin>>p>>t>>g1>>g2>>g3>>gj){ if(cha(g1,g2)<=t) { score=(g1+g2)/2; } else if(cha(g1,g3)>t && cha(g2,g3)>t){ score=gj; } else if(cha(g1,g3)<=t && cha(g2,g3)<=t){ score=maximum(g1,g2,g3); } else if(cha(g1,g3)<cha(g2,g3)){ score=(g1+g3)/2; } else score=(g2+g3)/2; cout<<fixed<<setprecision(1)<<score<<endl; } return 0; }
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