题目1002：Grading

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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

来源：

2011年浙江大学计算机及软件工程研究生机试真题

#include<stdio.h>#include<math.h>  int main(){    int  P,T,G1,G2,G3,GJ,a,b,c;    float grade;    while(scanf("%d %d %d %d %d %d",&P,&T,&G1,&G2,&G3,&GJ)!=EOF)    {        a=abs(G1-G2),b=abs(G1-G3),c=abs(G2-G3);//求绝对值        if(a<=T)            grade=(float)(G1+G2)/2;        else        {            if(b<=T&&c<=T)//求最大值            {                if(G1<G2)                    G1=G2;                if(G1<G3)                    G1=G3;                grade=(float)G1;            }            if(b>T&&c>T)                grade=(float)GJ;            if(b<=T)                grade=(float)(G1+G3)/2;            if(c<=T)                grade=(float)(G2+G3)/2;        }        printf("%.1f\n",grade);              }    return 0;      }/**************************************************************    Problem: 1002    Language: C    Result: Accepted    Time:0 ms    Memory:912 kb****************************************************************/