如何获取描述C语言中提到的输出

Question

编辑：

我使用scanf函数从控制台获得3个不同的数字。 （限制：只有整数可能是负数，正数）。我必须按照这样的顺序排列这些数字，以便它显示具有适当运算符关系的下降顺序，如：

Input:
    Enter 1st Num: 6
    Enter 2nd Num: 3
    Enter 3rd Num: 4

Expected Output: 6 > 3 > 2

＃

Input:
    Enter 1st Num: 3
    Enter 2nd Num: 2
    Enter 3rd Num: 6

Expected Output: 6 > 3 > 2

＃

Input:
    Enter 1st Num: 6
    Enter 2nd Num: 6
    Enter 3rd Num: 4

Expected Output: 6 = 6 > 4

＃

Input:
    Enter 1st Num: 6
    Enter 2nd Num: 3
    Enter 3rd Num: 3

Expected Output: 6 > 3 = 3

＃

Input:
    Enter 1st Num: 6
    Enter 2nd Num: 6
    Enter 3rd Num: 6

Expected Output: 6 = 6 = 6

＃

注意：我为2个数字做了同样的程序：

#include <stdio.h>
int main(void) {
    int a,b;
    printf("Enter 1st Num:");
    scanf("%d",&a);
    printf("Enter 2nd Num:");
    scanf("%d",&b);
    if (a>b)
    {
        printf("%d > %d",a,b);
    }
    else 
    {
        if (b > a){
            printf("%d > %d",b,a);
        }
        else{
            printf("%d = %d", a,b); 
        }
}
return 0;

}

以下代码的输出是：

Input:
    Enter 1st Num: 4
    Enter 2nd Num: 5

Output: 5 > 4

＃

Input:
    Enter 1st Num: 5
    Enter 2nd Num: 4

Output: 5 > 4

＃

Input:
    Enter 1st Num: 4
    Enter 2nd Num: 4

Output: 4 = 4

Answer 1

打印数字的方法：

void sort(int a, int b, int c)
{
   int top, mid, bot;
   if (a > b)
   {
      if (c > a)
      {
         top = c;
         mid = a;
         bot = b;
      }
      else
      {
         top = a;
         if (c > b)
         {
            mid = c;
            bot = b;
         }
         else
         {
            mid = b;
            bot = c;   
         }
      }
   }
   else
   {
      if (c > b)
      {
         top = c;
         mid = b;
         bot = a;
      }
      else
      {
         top = b;
         if (c > a)
         {
            mid = c;
            bot = a;
         }
         else
         {
            mid = a;
            bot = c;   
         }
      }
   }

   printf("%i %c %i %c %i \n", 
          top, (top == mid) ? '=' : '>',
          mid, (mid == bot) ? '=' : '>', 
          bot);
}