1 #include2 #include 3 #include 4 #include 5 #include 6 #include 7 using namespace std; 8 int map[15][15]; 9 int main(){10 int P,T,G1,G2,G3,GJ;11 while(cin>>P>>T>>G1>>G2>>G3>>GJ){12 if(abs(G1-G2)<=T){13 printf("%.1lf\n",1.0*(G1+G2)/2);14 }15 else{16 int m1=abs(G1-G3);17 int m2=abs(G3-G2);18 if(m1<=T&&m2<=T){19 int max=G1>G2?G1:G2;20 max=max>G3?max:G3;21 printf("%.1lf\n",max*1.0);22 }23 else{24 if(m1<=T||m2<=T){25 if(m1>m2){26 printf("%.1lf\n",(G3+G2)*1.0/2);27 }28 else{29 printf("%.1lf\n",(G3+G1)*1.0/2);30 }31 }32 else{33 printf("%.1lf\n",GJ*1.0);34 }35 }36 }37 }38 return 0;39 }
时间限制：1 秒
内存限制：32 兆
特殊判题：否
提交：16128
解决：4162
- 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
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- 分析：
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