Grading

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
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstdlib>
 4 #include <cmath>
 5 #include <cstring>
 6 #include <cctype>
 7 
 8 #include <vector>
 9 #include <list>
10 #include <deque>
11 #include <string>
12 #include <algorithm>
13 #include <stack>
14 #include <queue>
15 #include <map>
16 #include <set>
17 
18 using namespace std;
19 
20 
21 
22 
23 
24 
25 int main()
26 {
27 
28 
29 int p,t,g1,g2,g3,gj;
30 
31 while(scanf("%d%d%d%d%d%d",&p,&t,&g1,&g2,&g3,&gj)!=EOF)
32 {
33     int i,j,k;
34 
35 
36     int der=abs(g1-g2);
37 
38 
39     if(der<=t)
40     {
41         double ans=(g1+g2)*1.0/2;
42 
43         printf("%.1lf\n",ans);
44     }
45     else
46     {
47 
48         int der1=abs(g1-g3);
49         int der2=abs(g2-g3);
50 
51         if((der1<=t)&&(der2<=t))
52         {
53             int maxg=-1;
54             if(g1>maxg)
55             {maxg=g1;}
56             if(g2>maxg)
57                 maxg=g2;
58             if(g3>maxg)
59                 maxg=g3;
60 
61             double ans=maxg;
62 
63             printf("%.1lf\n",ans);
64 
65         }
66         else if(der1<=t)
67         {
68             double ans=(g1+g3)*1.0/2;
69 
70             printf("%.1lf\n",ans);
71 
72         }
73         else if(der2<=t)
74         {
75             double ans=(g2+g3)*1.0/2;
76 
77             printf("%.1lf\n",ans);
78 
79         }
80         else
81         {
82             double ans=gj;
83 
84         printf("%.1lf\n",ans);
85         }
86     }
87 }
88 
89 
90 
91     return 0;
92 }

 



posted @ 2012-05-30 20:07  cseriscser  阅读(350)  评论(0)  编辑  收藏  举报