【九度OJ】1002：Grading

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

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

源码：

#include

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

float ABS( float x ){   //求绝对值
    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给的分数和g2给的分数在可接受范围
            grade = (g1+g2)/2;
            printf( "%.1f\n", grade );
            continue;
        }
        else if( g3 - g1 <= t && g3 - g1 >= -t && g3 - g2 <= t && g3 - g2 >= -t ){  //g3给的分数和g1g2给的分数均在可接受范围内
            float maxmum = g1 > g2 ? g1 : g2;
            maxmum = maxmum > g3 ? maxmum : g3;    //找出三个中分数最大的那个

            grade = maxmum;
            printf( "%.1f\n", maxmum );
            continue;
        }
        else if( (g3 - g1 <= t && g3 - g1 >= -t) || (g3 - g2 <= t && g3 - g2 >= -t) ){   //g3给的分数和g1或者g2给的分数在可接受范围内（去除同时可在接受范围内的情况）
            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给的分数和g1或者g2给的均分数在可接受范围内
            grade = gj;
            printf( "%.1f\n", grade );
            continue;
        }
    }
}
/**************************************************************
    Problem: 1002
    User: 螺小旋
    Language: C++
    Result: Accepted
    Time:0 ms
    Memory:1020 kb
****************************************************************/