題目描述:
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.
- 輸入:
-
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].
- 輸出:
-
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.
- 樣例輸入:
-
20 2 15 13 10 18
- 樣例輸出:
-
14.0
- 來源:
#include<stdio.h>
#include<math.h>
int main()
{
int P,T,G1,G2,G3,GJ,a,b,c;
float grade;
while(scanf("%d %d %d %d %d %d",&P,&T,&G1,&G2,&G3,&GJ)!=EOF)
{
a=abs(G1-G2),b=abs(G1-G3),c=abs(G2-G3);//求絕對值
if(a<=T)
grade=(float)(G1+G2)/2;
else
{
if(b<=T&&c<=T)//求最大值
{
if(G1<G2)
G1=G2;
if(G1<G3)
G1=G3;
grade=(float)G1;
}
if(b>T&&c>T)
grade=(float)GJ;
if(b<=T)
grade=(float)(G1+G3)/2;
if(c<=T)
grade=(float)(G2+G3)/2;
}
printf("%.1f\n",grade);
}
return 0;
}
/**************************************************************
Problem: 1002
Language: C
Result: Accepted
Time:0 ms
Memory:912 kb
****************************************************************/