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

来源：

2011年浙江大学计算机及软件工程研究生机试真题

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思路：

• 简单选择判断即可～

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代码：

 1 #include<stdio.h> 2 int jud(int x,int y); 3 int t; 4 int main(int argc, char const *argv[]) 5 { 6     int p,g1,g2,g3,gj; 7     double sco; 8     while(scanf("%d %d %d %d %d %d",&p,&t,&g1,&g2,&g3,&gj)==6) 9     {10         if(jud(g1,g2))11             sco=(g1+g2)/2.0;12         else if(jud(g1,g3)&&!jud(g2,g3))13             sco=(g1+g3)/2.0;14         else if(jud(g2,g3)&&!jud(g1,g3))15             sco=(g2+g3)/2.0;16         else if(jud(g1,g3)&&jud(g2,g3))17             {18                 sco=g1>g2 ? g1:g2;19                 sco=sco>g3 ? sco:g3;20             }21         else if(!jud(g1,g2)&&!jud(g2,g3))22             sco=gj;    23         printf("%.1f\n",sco);24     }25 26     return 0;27 }28 29 int jud(int x,int y)30 {31     if(x-y<=t&&x-y>=-t)32         return(1);33     else34         return(0);35 }