Nine degrees OJ 1002 Grading

Topic 1002:grading

time limit:1 seconds

Memory limit:32 MB

Special question: No

submitted:15686

Resolution:4053

Title 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-to-assign each exam problem to 3 independent experts. If they do and agree to each other, a judge are invited to make the final decision. Now is asked to the 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 would first be assigned to 2 experts, to obtain G1 and G2. If the difference is within the tolerance, which is, if | g1-g2| ≤t, this problem ' s grade would be the average of G1 and G2.
? If the difference exceeds T, the 3rd expert would give G3.
? If G3 is within the tolerance with either G1 or G2, but not both, then this problem ' s grade would 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'll be the maximum of the three grades.
? If G3 is within the tolerance with neither G1 nor G2, a judge would give the final grade GJ.

input :

    each input file could contain more than  One test case.
    occupies a line containing six positive integers:p, T, G1, G2, G3, and GJ, as described In the problem. It is guaranteed and the grades is valid, which is, in the interval [0, P].

Output:

For each test case you should output the final grade of the problem in a line. The answer must is 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", &AMP;P,&AMP;T,&AMP;G1,&AMP;G2,&AMP;G3,&AMP;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 Resu lt:accepted time:0 Ms memory:912 kb****************************************************************/ 

Nine degrees OJ 1002 Grading