Foreword:
21. Regardless of whether you can enter the retest or not, record the garbage code written on the road. I originally gnawed on "Algorithm Notes", but I felt too much to do it, so I changed it to the Kingway Computer Test Guide.
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 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.
Enter description
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].
Output description:
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.
answer:
#include<stdio.h>
#include<cmath>
int main()
{
double P, T, G1, G2, G3, GJ;
while (scanf("%lf%lf%lf%lf%lf%lf", &P, &T, &G1, &G2, &G3, &GJ) != EOF) {
if (fabs(G1 - G2) <= T)
GJ = (G1 + G2) / 2;
else {
if ((fabs(G3 - G2) <= T) && fabs(G3 - G2) <= T)
GJ = fmax(fmax(G1, G2), G3);
else
if ((fabs(G3 - G2) > T) && fabs(G3 - G2) > T);
else {
if (G3 - G1 > G3 - G2)
GJ = (G3 + G2) / 2;
else
GJ = (G3 + G1) / 2;
}
}
printf("%.1lf\n", GJ);
}
return 0;
}