【菜鸟提问】一道onlinejudge上面的题目，百思不得骑姐。。。。。
本帖最后由 zhjxwuhan 于 2012-10-03 18:14:30 编辑 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
我的解法为：
#include <stdio.h>
#include <stdlib.h>
double max(double,double,double);
int main(void)
{
double p,t,g1,g2,g3,gj;
while(scanf("%lf %lf %lf %lf %lf %lf",&p,&t,&g1,&g2,&g3,&gj)!=EOF)
{
if(abs(g1-g2)<=t)
{
printf("%.1f\n",(g1+g2)/2);
return 2;
}
int s1,s2;
s1=(int)abs(g3-g1)-t;
s2=(int)abs(g3-g2)-t;
if(s1>0)
{
if(s2>0)
{
printf("%.1f\n",gj);
return 0;
}
if(s2<=0)
{
printf("%.1f\n",(g3+g2)/2);
return 0;
}
}
else
{
if(s2>0)
{
printf("%.1f\n",(g3+g1)/2);
return 0;
}
else
{
printf("%.1f\n",max(g1,g2,g3));
return 0;
}
}
}
return 1;
}
double max(double x,double y,double z)
{
double temp;
temp=x>y?x:y;
temp=temp>z?temp:z;
return temp;
}
#include <stdio.h>
#include <stdlib.h>
double max(double,double,double);
int main(void)
{
double p,t,g1,g2,g3,gj;
while(scanf("%lf %lf %lf %lf %lf %lf",&p,&t,&g1,&g2,&g3,&gj)!=EOF)
{
if(abs(g1-g2)<=t)
{
printf("%.1f\n",(g1+g2)/2);
}
else{
int s1,s2;
s1=(int)abs(g3-g1)-t;
s2=(int)abs(g3-g2)-t;
if(s1>0)
{
if(s2>0)
{
printf("%.1f\n",gj);
}
else if(s2<=0)
{
printf("%.1f\n",(g3+g2)/2);
}
}
else
{
if(s2>0)
{
printf("%.1f\n",(g3+g1)/2);
}
else
{
printf("%.1f\n",max(g1,g2,g3));
}
}
}
}
return 0;
}
double max(double x,double y,double z)
{
double temp;
temp=x>y?x:y;
temp=temp>z?temp:z;
return temp;
}
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
double max(double,double,double);
int main(void)
{
double p,t,g1,g2,g3,gj;
while(scanf("%lf %lf %lf %lf %lf %lf",&p,&t,&g1,&g2,&g3,&gj)!=EOF)
{
if(abs(g1-g2)<=t)
{
printf("%.1f\n",(g1+g2)/2);
continue; //有多组测试，不能再做完一个就返回了，应该继续做下一个
}
double s1,s2;
s1=abs(g3-g1)-t; //要用double型，否则会有截断误差的，如果s1=0.5的话，用int就变成0了
s2=abs(g3-g2)-t;
if(s1>0)
{
if(s2>0)
{
printf("%.1f\n",gj);
}
if(s2<=0)
{
printf("%.1f\n",(g3+g2)/2);
}
}
else
{
if(s2>0)
{
printf("%.1f\n",(g3+g1)/2);
}
else
{
printf("%.1f\n",max(g1,g2,g3));
}
}
}
return 1;
}
double max(double x,double y,double z)
{
double temp;
temp=x>y?x:y;
temp=temp>z?temp:z;
return temp;
}
