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;}