【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(<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.

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<stdio.h>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****************************************************************/