九度OJ：1002-Grading

时间限制：1 秒内存限制：32 兆特殊判题：否提交：24102解决：6126

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

翻译题目:<忽略丑丑的字和翻译

代码如下：

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

int main()
{
	int p=0,t=0,g1=0,g2=0,g3=0,g4=0;
	int a,b,max;
	float g;
	while(scanf("%d %d %d %d %d %d",&p,&t,&g1,&g2,&g3,&g4)!=EOF){
		if(abs(g1-g2) <= t){
			g=(float)(g1+g2)/2;
		}
		else if(abs(g1-g3)<=t || abs(g2-g3)<=t)
		{
			a=abs(g1-g3);
			b=abs(g2-g3);
			if(a<t && b<t)
			{
				max=g1;
				if(g2>max){
					max=g2;
				}
				if(g3>max){
					max=g3;
				}
				g=(float)max;
			}
			else if(a<b){
				g=(float)(g1+g3)/2;
			}
			else{
				g=(float)(g2+g3)/2;
			}
		}
		else{
			g=(float)g4;
		}
		printf("%.1f\n",g);
	}
# 	return 0;
}`

结果：

• 输出文件名: D:\DevC++\程序\two.exe
• 输出大小: 362.3271484375 KiB
• 编译时间: 0.45s

注意点：输入是有多组数据的。