1002_Grading

// 1002_Grading.cpp : 定義控制檯應用程序的入口點。
//題目1002：Grading
//時間限制：1 秒內存限制：32 兆特殊判題：否提交：21415解決：5522
//題目描述：
//    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.
//輸入：
//    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].
//輸出：
//    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.
//樣例輸入：
//20 2 15 13 10 18
//樣例輸出：
//14.0
//來源：
//2011年浙江大學計算機及軟件工程研究生機試真題

#include "stdafx.h"
#include "stdio.h"
#include "math.h"

int main()
{
    int p,t,g1,g2,g3,gj;
    while(scanf("%d %d %d %d %d %d",&p,&t,&g1,&g2,&g3,&gj)!=EOF)
    {
        if(abs(g1-g2)<=t)
            printf("%.1f\n",(float)(g1+g2)/2);
        else if(abs(g3-g1)<=t && abs(g3-g2)<=t)
        {
            int max = g1;
            if(g2>g1) max =g2;
            if(g3>g1&&g3>g2) max= g3;
            printf("%.1f\n",(float)max);
        }
        else if(abs(g3-g1)<=t)
            printf("%.1f\n",(float)(g3+g1)/2);
        else if (abs(g3-g2)<=t)
            printf("%.1f\n",(float)(g3+g2)/2);
        else if(abs(g3-g2)>t &&abs(g3-g1)>t)
            printf("%.1f\n",(float)gj);
    }
    //printf("%.1f",(float)(3+4)/2);
    return 0;
}

/*

abs();

*/