九度OJ 题目1002

题目1002：Grading

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：15686

解决：4053

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

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#include<stdio.h>  

#include<stdlib.h>  

#include<math.h>  

int P,T,G1,G2,G3,GJ;  

int my_max(int x,int y)  

{  

    return x>y?x:y;  

}  

int main(int argc, char *argv[])  

{  

    while(scanf("%d %d %d %d %d %d",&P,&T,&G1,&G2,&G3,&GJ)!=EOF)  

    {  

        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);  

    }  

    return 0;  

}  

   

/************************************************************** 

    Problem: 1002 

    User: kirchhoff 

    Language: C 

    Result: Accepted 

    Time:0 ms 

    Memory:912 kb 

****************************************************************/