题目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

AC代码：

//本题不难，看清题目分情况讨论就行，但有几个地方需要注意;//1.abs()在头文件stdlib.h里，fabs()在math.h里//2.读入的是整数，输出的是格式化的浮点数#include<stdio.h>#include<math.h>int main() {//freopen("in.txt","r",stdin);int P, T, G1, G2, G3, GJ;double grade;while(scanf("%d %d %d %d %d %d",&P,&T,&G1,&G2,&G3,&GJ)!=EOF) {if((int)fabs(G1-G2)<=T)grade = (G1+G2)/2.0; //注意除数是2.0,不然'/'就成取整符号而不是除号了else {int t1, t2;t1 = (int) fabs(G3-G1); //fabs的返回值的double，要强制转换为intt2 = (int) fabs(G3-G2);if((t1<=T && t2>T) || (t2<=T && t1>T)) {if(t1 < t2)grade = (G3+G1)/2.0;elsegrade = (G3+G2)/2.0;}if(t1<=T && t2<=T) {int max;max = G1 > G2 ? G1 : G2;max = max > G3 ? max : G3;grade = max;}if(t1>T && t2>T)grade = GJ;}printf("%.1lf\n",grade); //注意输出格式}return 0;}