9-degree OJ 1002 Grading, 9-degree oj1002grading

9-degree OJ 1002 Grading, 9-degree oj1002grading

Question 1002: Grading

Time Limit: 1 second

Memory limit: 32 MB

Special question: No

Submit: 15686

Solution: 4053

Description:

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 demo. 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.

Input:

Each input file may contain in 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].

Output:

For each test case you shoshould output the final grade of the problem in a line. The answer must be accurate to 1 decimal place.

Sample input:

20 2 15 13 10 18

Sample output:

14.0

#include<stdio.h>#include<stdlib.h>#include<math.h>int P,T,G1,G2,G3,GJ;int my_max(int x,int y){    return x>y?x:y;}int main(int argc, char *argv[]){    while(scanf("%d %d %d %d %d %d",&P,&T,&G1,&G2,&G3,&GJ)!=EOF)    {        if(abs(G1-G2)<=T)            printf("%.1lf\n",(double)(G1+G2)/2.0);        else if(abs(G1-G3)<=T&&abs(G2-G3)<=T)            printf("%.1lf\n",my_max(my_max(G1,G2), G3));        else if(abs(G1-G3)>T&&abs(G2-G3)>T)            printf("%.1lf\n",(double)GJ);        else if(abs(G1-G3)<=T&&abs(G2-G3)>T)            printf("%.1lf\n",((double)(G3+G1))/2.0);        else if(abs(G2-G3)<=T&&abs(G1-G3)>T)            printf("%.1lf\n",((double)(G3+G2))/2.0);    }    return 0;} /**************************************************************    Problem: 1002    User: kirchhoff    Language: C    Result: Accepted    Time:0 ms    Memory:912 kb****************************************************************/