题目1002：Grading

题目1002：Grading

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：24306

解决：6174

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

答疑：

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#include <stdio.h>#include <iostream>#include <stack>#include <string.h>#include <queue>#include <cmath>#include <vector>#include <algorithm>#include <map>#include <set>#include <string>using namespace std;typedef long long LL;int ans[1000000]; int main() {    //freopen("in.txt", "r", stdin);    //freopen("out.txt", "w", stdout);    double P, T, G1, G2, G3, GJ;    while(scanf("%lf %lf %lf %lf %lf %lf", &P, &T, &G1, &G2, &G3, &GJ) != EOF) {        if(abs(G1 - G2) <= T){            printf("%.1lf\n", (G1 + G2) / 2);        }else if((abs(G3 - G1) <= T && abs(G3 - G2) > T) || (abs(G3 - G2) <= T && abs(G3 - G1) > T)){            if(abs(G3 - G1) < abs(G3 - G2)){                printf("%.1lf\n", (G3 + G1) / 2);            }else{                printf("%.1lf\n", (G3 + G2) / 2);            }        }else if((abs(G3 - G1) <= T && abs(G3 - G2) <= T)){            double temp = G3 > G1 ? G3 : G1;            temp = temp > G2 ? temp : G2;            printf("%.1lf\n", temp);        }else if((abs(G3 - G1) > T && abs(G3 - G2) > T)){            printf("%.1lf\n", GJ);        }    }       return 0;} /**************************************************************    Problem: 1002    User: Crazy_man    Language: C++    Result: Accepted    Time:0 ms    Memory:5424 kb****************************************************************/