【九度OJ】1002：Grading

地址：
http://ac.jobdu.com/problem.php?pid=1002
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.

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

源碼：

#include<stdio.h>

float p, t, g1, g2, g3, gj;
float grade;

float ABS( float x ){   //求絕對值
    if( x > 0 ){
        return x;
    }
    else {
        return -x;
    }
}
int main(){
    while( scanf( "%f %f %f %f %f %f", &p, &t, &g1, &g2, &g3, &gj) != EOF ){
        if( g1 - g2 <= t && g1 - g2 >= -t ){    //g1給的分數和g2給的分數在可接受範圍
            grade = (g1+g2)/2;
            printf( "%.1f\n", grade );
            continue;
        }
        else if( g3 - g1 <= t && g3 - g1 >= -t && g3 - g2 <= t && g3 - g2 >= -t ){  //g3給的分數和g1g2給的分數均在可接受範圍內
            float maxmum = g1 > g2 ? g1 : g2;
            maxmum = maxmum > g3 ? maxmum : g3;    //找出三個中分數最大的那個

            grade = maxmum;
            printf( "%.1f\n", maxmum );
            continue;
        }
        else if( (g3 - g1 <= t && g3 - g1 >= -t) || (g3 - g2 <= t && g3 - g2 >= -t) ){   //g3給的分數和g1或者g2給的分數在可接受範圍內（去除同時可在接受範圍內的情況）
            float g3g1 = ABS( g3 - g1 );
            float g3g2 = ABS( g3 - g2 );

            if( g3g1 > g3g2 ){
                grade = (g3+g2)/2;
                printf( "%.1f\n", grade );
                continue;
            }
            else{
                grade = (g3+g1)/2;
                printf( "%.1f\n", grade );
                continue;
            }
        }
        else{    //g3給的分數和g1或者g2給的均分數在可接受範圍內
            grade = gj;
            printf( "%.1f\n", grade );
            continue;
        }
    }
}
/**************************************************************
    Problem: 1002
    User: 螺小旋
    Language: C++
    Result: Accepted
    Time:0 ms
    Memory:1020 kb
****************************************************************/