1002 Grading(2011年浙江大学计算机及软件工程研究生机试真题)

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

//Grading#include<iostream>#include<cstdlib> #include<iomanip>using namespace std;int qmax(int x,int y){return x<y?x:y;}int main(){int p,t,g1,g2,g3,g;double x;while(cin>>p>>t>>g1>>g2>>g3>>g){if(abs(g1-g2)<=t){x=(g1+g2)/2.0;cout<<setprecision(1)<<setiosflags(ios::fixed)<<x<<endl;continue;}else if(abs(g1-g3)>t&&abs(g2-g3)>t){x=g;  cout<<setprecision(1)<<setiosflags(ios::fixed)<<x<<endl;  continue;}else if((abs(g1-g3)>t&&abs(g2-g3)<=t)||(abs(g1-g3)<=t&&abs(g2-g3)>t)){    if(qmax(abs(g1-g3),abs(g2-g3))==abs(g1-g3))    x=(g1+g3)/2.0;    else         x=(g2+g3)/2.0;        cout<<setprecision(1)<<setiosflags(ios::fixed)<<x<<endl;        continue;}else {    if(g1>=g2&&g1>=g3)    x=g1;    else if(g2>=g1&&g2>=g3)    x=g2;    else     x=g3;    cout<<setprecision(1)<<setiosflags(ios::fixed)<<x<<endl;        continue;}}return 0;}