题目1002：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

来源：2011年浙江大学计算机及软件工程研究生机试真题

#include<stdio.h>
#include<stdlib.h>
int abs(int a){
return a>0?a:a*(-1);
}
int main(){
int P,T,G1,G2,G3,GJ;
float grade,temp;
while(scanf("%d %d %d %d %d %d",&P,&T,&G1,&G2,&G3,&GJ)!=EOF){
if(abs(G1-G2)<=T) grade=(G1+G2)/2.0;
else{
if(abs(G1-G3)<=T && abs(G2-G3)<=T){
temp=G1>G2?G1:G2;
grade=temp>G3?temp:G3;
}
else if(abs(G1-G3)<=T || abs(G2-G3)<=T){
temp=(float)(abs(G1-G3)<abs(G2-G3)?G1:G2);
grade=(float)(G3+temp)/2.0;
}
else grade=(float)GJ;
}
printf("%.1f\n",grade);
}
// getchar();
return 0;
}