2011年浙江大学计算机及软件工程研究生机试真题（2）

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：

#include<stdio.h>
#include <stdlib.h>
//using namespace std;

int main(){
	int p,t,g1,g2,g3,gj;
	float max;

	while(scanf("%d%d%d%d%d%d",&p,&t,&g1,&g2,&g3,&gj)!=EOF){
		if(abs(g1-g2)<=t){
			max=(float)(g1+g2)/2;
			printf("%.1f\n",max);
			continue;
		}
		else if(abs(g3-g1)<=t&&abs(g3-g2)<=t)
		{
			g3=g3>g1?g3:g1;
			g3=g3>g2?g3:g2;
			max=(float)g3;
			printf("%.1f\n",max);
			continue;
		}
		else if(abs(g3-g1)<=t||abs(g3-g2)<=t){
			max=(float)((abs(g3-g1)>abs(g3-g2)?g2:g1)+g3)/2;
			printf("%.1f\n",max);
			continue;
		}
		else {
			max=(float)gj;
			printf("%.1f\n",max);
			continue;
		}
	}
	return 0;
}

感觉代码写的很乱，一开始就搞错了一步，虽然给出的例子能运行，但是提交老是WA，但一直没有发现这个错误，一直改来改去，最后认真验证数据时才发现问题所在，懒得去优化了，就这样看吧！