九度题目1002：Grading

May 16, 2016
作者：dengshuai_super
出处：http://blog.csdn.net/dengshuai_super/article/details/51425105
声明：自由转载，转载请注明作者及出处。

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题目1002：Grading

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

题目分析：

输入为6个参数：P, T, G1, G2, G3, and GJ ，全部的成绩都满足在[0, P]区间内。

分级规则：

•问题将首先被分配到2个专家，得到G1和G2。如果差异是可以接受的，也就是满足|G1 - G2| ≤ T，这一问题的成绩将取G1和G2的平均值。

•如果差异超过T，第三专家将给G3。

•如果G3和G1或G2的差值只有其中一个在公差范围内，那么这个问题的成绩将取G3和最接近的成绩的平均值。

•如果G3和G1或G2的差值都在公差范围内，那么这一问题的成绩将取G1，G2，G3的最大值。

•如果没有G3和G1 或G2的差值都不在的公差范围内，将作出最后的等级GJ。

代码如下：

/******************************************************************************   九度题目1002：Grading *******************************************************************************   by Deng shuai May 16 2016*   http://blog.csdn.net/dengshuai_super*******************************************************************************   Copyright (c) 2016, Deng Shuai*   All rights reserved.*****************************************************************************/

#include<stdio.h>#include<stdlib.h>#include<math.h>//#include<time.h>int P,T,G1,G2,G3,GJ;double my_max(double x,double y){    return x>y?x:y;}int main(int argc, char *argv[]){    //clock_t start, end;    //start = clock(); //记录开始时间    while(scanf("%d %d %d %d %d %d",&P,&T,&G1,&G2,&G3,&GJ)!=EOF)    {        if((G1>P||G1<0)||(G2>P||G2<0)||(G3>P||G3<0)||(GJ>P||GJ<0))         {              break;           }        if(abs(G1-G2)<=T)            printf("%.1lf\n",(double)(G1+G2)/2.0);        else if(abs(G1-G3)<=T&&abs(G2-G3)<=T)            printf("%.1lf\n",my_max(my_max(G1,G2), G3));        else if(abs(G1-G3)>T&&abs(G2-G3)>T)            printf("%.1lf\n",(double)GJ);        else if(abs(G1-G3)<=T&&abs(G2-G3)>T)            printf("%.1lf\n",((double)(G3+G1))/2.0);        else if(abs(G2-G3)<=T&&abs(G1-G3)>T)            printf("%.1lf\n",((double)(G3+G2))/2.0);       //end = clock(); //结束时间       //printf("spend %lf s",(double)(end - start) / CLOCKS_PER_SEC);    }    return 0;}

截图：