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
#include<iostream>#include<cstdio>using namespace std;double abs(double a,double b){if(a-b<0){return b-a;}else{return a-b;}}int main(){int P=0,T=0;while(scanf("%d",&P)!=EOF){cin>>T;double temp=0;double G1=0,G2=0,G3=0,GJ=0;cin>>G1>>G2>>G3>>GJ;if((G1-G2)<=T&&(G1-G2)>=-T){temp=(G1+G2)/2;printf("%.1lf\n",temp);//cout<<temp<<endl;continue;}if((G1-G3)<=T&&(G1-G3)>=-T&&(G3-G2)<=T&&(G3-G2)>=-T){int max=G1;if(max<G2) max=G2;if(max<G3) max=G3;printf("%.1lf\n",max);//cout<<max<<endl;}else if(((G1-G3)<=T&&(G1-G3)>=-T)||((G3-G2)<=T&&(G3-G2)>=-T)){if(abs(G1,G3)<abs(G2,G3)){ temp=(G1+G3)/2;}else{temp=(G2+G3)/2;}printf("%.1lf\n",temp);//cout<<temp<<endl;}else{printf("%.1lf\n",GJ);请保留两位小数，如果是double,则用lf,如果是float,则用f//cout<<GJ<<endl;}}return 0;}