OJ_1002

#include <iostream>#include <iomanip>using namespace std;int getfab(int a,int b){    if(a>b)           return a-b;    else return b-a;}bool isTol(int a,int b,int t){     if(getfab(a,b)<=t)                       return true;     return false;}int getmax(int a,int b,int c){    int max;    if(a>b)    {         max=a;      }else    {         max=b;     }    if(max<c)             max=c;    return max;}int getclosest(int a,int b,int c){    int f1=    getfab(a,c);    int f2=getfab(b,c);    if(f1<f2)return a;    return b;}void func(){     int p,t,g1,g2,g3,gj;     while(cin>>p>>t>>g1>>g2>>g3>>gj)     {         double grade=0;         if(isTol(g1,g2,t))               grade=(double)(g1+g2)/2.0;         else         if(isTol(g3,g1,t)||isTol(g3,g2,t))         {               int closest=getclosest(g1,g2,g3);               if(isTol(g3,g1,t)&&isTol(g3,g2,t))                   grade=getmax(g1,g2,g3);               else                   grade=(double)(closest+g3)/2.0;         }         else{              grade=gj;         }                  cout<<fixed<<setprecision(1);         cout<<grade<<endl;                                                   }        }int main(int argc, char *argv[]){    //printf("Hello, world\n");func();return 0;}

难度在于看明白题目

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.

// 对每个问题，有一个满分P和一个容忍分T（T<P），成绩计算规则是：

// 问题首先给两个专家，得到G1 G2，如果G1G2在容忍分范围内即 |G1 - G2| ≤ T,这个问题的成绩是G1与G2的平均值

// 如果 |G1 - G2| >T，第三个专家给出G3

// 如果G3与G1,G2中的一个在容忍分范围内，问题成绩将是G3与G1,G2中最接近的一个的平均值

// 如果G3与G1,G2都在容忍分范围内，成绩将是三者中最大值

// 如果G3与G1,G2都不在容忍分范围内，最终给出GJ作为成绩

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