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1551 - Grading

通过次数

0

提交次数

0

时间限制 : 1 秒 内存限制 : 32 MB

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.

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题目输入

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 

<span style="font-size:12.0pt;font-family:&quot;color:#333333;background:white;">GJ, as described in the problem. It is guaranteed that all the grades are valid, that is, in the interval [0, P].</span>


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题目输出

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.


输入/输出样例

输入格式

20 2 15 13 10 18

输出格式

14.0

C语言解答

#include <stdio.h>

int abs(int a)
{
  if(a<0)
  {
    return -a;
  }
  return a;
}
float maxnum(int a,int b,int c)
{
  float max=(float)a;
  if(max<b)
  {
    max=b;
  }
  if(max<c)
  {
    max=c;
  }
  return max;
}
int main()
{
  while(1)
  {
    int p,t,g1,g2,g3,gj;
    if(scanf("%d%d%d%d%d%d",&p,&t,&g1,&g2,&g3,&gj)==EOF)
    {
      break;
    }
    if(abs(g1-g2)<=t)
    {
      printf("%.1f\n",(float)(g1+g2)/2);
      continue;
    }
    else
    {
      if(abs(g1-g3)<=t &&abs(g2-g3)<=t)
      {
        printf("%.1f\n",maxnum(g1,g2,g3));
        continue;
      }
      else if(abs(g1-g3)>t &&abs(g2-g3)<=t)
      {
        printf("%.1f\n",(float)(g2+g3)/2);
        continue;
      }
      else if(abs(g1-g3)<=t && abs(g2-g3)>t)
      {
        printf("%.1f\n",(float)(g1+g3)/2);
        continue;
      }
      else if(abs(g2-g3)>t&&abs(g1-g3)>t)
      {
        printf("%.1f\n",(float)gj);
        continue;
      }
    }
  }
}

C++解答

#include <stdio.h>
#include <math.h>

int main(){

	// 定义变量存储总分、差值以及四个分数
    double p, t, g1, g2, g3, gj;
    while(scanf("%lf", &p) != EOF){		// 如果没有读到文件结尾，说明还有测试数据

    	// 读入剩下的测试数据
        scanf("%lf%lf%lf%lf%lf", &t, &g1, &g2, &g3, &gj);
        if(fabs(g1-g2) <= t){					// 如果 g1 和 g2 的差值不大于 t，则输出 g1 和 g2 的平均分
            printf("%.1lf\n", (g1+g2)/2);
        }else{
            double tmp1 = fabs(g3-g1);			// 计算 g3 与 g1 的差值
            double tmp2 = fabs(g3-g2);			// 计算 g3 与 g2 的差值
            if(tmp1<=t && tmp2<=t){				// 如果两个差值均不大于 t，则输出三个分数中的最大值
                printf("%.1lf\n", g1>g2 ? (g1>g3?g1:g3) : (g2>g3?g2:g3));
            }else if(tmp1 <= t){				// 如果 g3 与 g1 的差值不大于t，则输出 g1 与 g3 的平均分
                printf("%.1lf\n", (g1+g3)/2);
            }else if(tmp2 <= t){				// 如果 g3 与 g2 的差值不大于t，则输出 g2 与 g3 的平均分
                printf("%.1lf\n", (g2+g3)/2);
            }else{								// 否则两个差值均大于t，则输出另外给出的分数 gj
                printf("%.1lf\n", gj);
            }
        }
    }

    return 0;
}
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