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

代码

#include <iostream>
#include <stdio.h>
//#include <cmath>
#include <cstdlib>

using namespace std;

int main()
{
int P,T,G1,G2,G3,GJ;
float GF;
while(scanf("%d %d %d %d %d %d",&P,&T,&G1,&G2,&G3,&GJ)!=EOF)
{
//cout << P << " "<<T<< " "<<G1<< " "<<G2<< " "<<G3<< " "<<GJ << endl;
if(abs(G1-G2) <= T)
{
GF = (float)(G1*1.0 + G2*1.0) /2;
printf("%.1lf\n",GF);
} else if(abs(G3-G2) <= T && abs(G3-G1) <= T)
{
float temp;
if(G1 > G2)
{
temp = G1;
if(G1 < G3)
temp = G3;
} else
{
temp = G2;
if(G2 < G3)
temp = G3;
}
printf("%.1lf\n",temp);
} else if(abs(G3-G2) <= T || abs(G3-G1) <= T)
{
if(abs(G3-G2) <= T)
{
GF = (float)(G3 + G2)/2;
printf("%.1lf\n",GF);
}
if(abs(G3-G1) <= T)
{
GF = (float)(G3 + G1)/2;
printf("%.1lf\n",GF);
}
} else
{
GF = (float)GJ;
printf("%.1lf\n",GF);
}
}
return 0;
}

注意：printf("%.1lf\n",GF);若GF为整数则输出零 需要强制类型转换