题目1002：Grading

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：16499

解决：4267

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

要注意是“读文件”（Each input file may contain more than one test case.），所以要有EOF，读完文件的标志。

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <math.h>using namespace std;int main(){    int p=20,t=2,g1=16,g2=13,g3=10,gj=18;    while(scanf("%d %d %d %d %d %d",&p,&t,&g1,&g2,&g3,&gj)!=EOF)    {        double grade;        int a = abs(g1-g2),b=abs(g3-g1),c=abs(g3-g2);        if(a<=t)            grade = (g1+g2)*1.0/2;        else        {            if(b<=t&&c<=t)            {                int max =g1;                if(g2>max) max = g2;                if(g3>max) max = g3;                grade = max*1.0;            }            else if(b<=t)                grade = (g1+g3)*1.0/2;            else if(c<=t)                grade = (g2+g3)*1.0/2;            else                grade = gj*1.0;        }        printf("%.1lf\n",grade);    }    return 0;}