C程序习题-输出学生成绩[10.15]

    有一个班4个学生，5门课。1.求第一门课的平均分。2.找出有两门以上课程不及格的学生，并输出他们的学号和全部课程成绩和平均成绩.3.找出平均成绩在90分以上，或者全部课程成绩在85分以上的学生。分别编写三个函数实现以上3个要求。

    我们可以把学生成绩保存在二维数组，通过函数传递过去数据，在函数中定义一维数组的指针变量。根据要求，对每个学生的成绩进行扫描，输出结果。代码如下：

 

#include <stdio.h>

float average(int [][5],int,int);
void search_fail2(int [][5],int);
void search_good(int [][5],int);
void print_source(int [][5],int);
int main(int argc, char *argv[])
{
    int arr[4][5] ={{80,50,70,65,30},{77,88,99,66,88},{98,95,96,96,92},{66,68,80,30,40}}; //test

    int i;
    /*
    printf("please input 5 number(4 line):\n");
    for (i = 0; i < 4;i++)
    {
        scanf("%d %d %d %d",&arr[i][0],&arr[i][1],&arr[i][2],&arr[i][3],&arr[i][4]);
    }
    */
    printf("the source data is :\n");
    print_source(arr,4);
    printf("\nthe 1 kemu average is :%.2f\n\n",average(arr,4,0));
    search_fail2(arr,4);
    search_good(arr,4);
    system("pause");
    return 0;
}

float average(int arr[][5],int n,int k)
{
      int i,j,(*p)[5];
      p = arr;
      float result = 0;
      for (i = 0; i < n; i++)
      {
          result += *(*(p+i)+k);
      }
      return result / n;
}

void search_fail2(int arr[][5],int n)
{
     int i,j,(*p)[5],count = 0;
     float average = 0;
     p = arr;
     for (i = 0; i < n;i++)
     {
         for (j = 0 ;j < 5; j++)
         {
             if (*(*(p+i)+j) < 60)
             {
                count++;
             }
         }
         
         if (count >= 2)
         {
            printf("NO.%d is fall,the data is \n",i);
            for (j = 0; j < 5; j++)
            {
                printf("%d ",*(*(p+i)+j));
                average += *(*(p+i)+j);
            }
            printf("\nthe average is %.2f\n\n",average / 5);
         }
         count =0;
         average = 0;
         
     }
}

void print_source(int arr[][5],int n)
{
     int i,j;
     int (*p)[5] = arr;
     for (i = 0; i < n; i++)
     {
         printf("NO.%d: ",i);
         for (j = 0; j < 5;j++)
         {
             printf("%d ",*(*(p+i)+j));
         }
         printf("\n");
     }
}

void search_good(int arr[][5],int n)
{
     int i,j,count = 0;
     int (*p)[5] = arr;
     float average = 0;
     for (i = 0; i < n ; i++)
     {
         for (j = 0; j < 5; j++)
         {
             if (*(*(p + i) + j) >= 85)
             {
                count++;
             }
             average += *(*(p + i) + j);
         }
         if (count == 5 || (average / 5 > 90) )
         {
            printf("the good student is NO.%d\n",i);
         }
         count = 0;
         average = 0;
         
     }
}


运行结果如下：