九度OJ 题目1002：Grading

一题目描述：

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

二.题目分析

简单的模拟过程，注意最后结果要求输出精度为1，由于涉及到除法运算，数据类型最好为double型.

三.代码

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

double max(double x,double y,double z)
{
double m=-1;
if(x>m)
m=x;
if(y>m)
m=y;
if(z>m)
m=z;
return m;
}
int main()
{
double P,T,G1,G2,G3,GJ,ans,close;
int i,j;

freopen("1002.txt","r",stdin);

while(scanf("%lf%lf%lf%lf%lf%lf",&P,&T,&G1,&G2,&G3,&GJ)!=EOF)
{

if(fabs(G1-G2)<=T)
ans=(G1+G2)/2;
else if((fabs(G1-G3)<=T)&&(fabs(G2-G3)<=T))
ans=max(G1,G2,G3);
else if((fabs(G1-G3)<=T)||(fabs(G2-G3)<=T))
{
if((fabs(G1-G3))<(fabs(G2-G3)))
close=G1;
else
close=G2;

ans=(G3+close)/2;
}
else
ans=GJ;

printf("%.1f\n",ans);
}
return 0;
}