【菜鸟提问】一道onlinejudge上面的题目，百思不得骑姐。。。。。
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：

C/C++ code

#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;}

哪里有错有错，一直是WRONG ANSWER
题目地址为：http://ac.jobdu.com/problem.php?pid=1002

[解决办法]
改了下你的代码，如下可以AC

C/C++ code

#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;} 
 
[解决办法]
有两点错误，在注释里。
C/C++ code
#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;}