案例代码资料文件与下载

//案例7.1*************************************

#include "stdio.h"
int main()
{
     int c;
     float t,r;  

     printf("请输入订单总额(万元):");
     scanf("%f",&t);
     c=(int)(t/5);
     switch(c)
     {

         case 0:
               r=t*0.2;break;
         case 1:    
               r=5*0.2+(t-5)*0.15 ;break;
         case 2:
         case 3:
               r=5*0.2+(t-5)*0.15;break;
         case 4:
         case 5:
         case 6:
         case 7:
         case 8:
         case 9:    
               r=5*0.2+5*0.15+10*0.1+(t-20)*0.08;break;
         default: 
               r=5*0.2+5*0.15+10*0.1+30*0.08+(t-50)*0.05;
     }

     //输出结果
     printf("订单总额为%5.2f万元的毛利润是%5.2f万元\n",t,r);
     return 0;
}

//案例7.2*************************************

#include "stdio.h"

int main()
{
     int c;
     float t, r , d ;  
     printf("请输入订单总额(万元):");
     scanf("%f",&t);

     d=t;r=0;
     if(t>50)
          c=10; 
     else
          c=(int)(t/5);
     switch(c)
     { 
          case 10: 
                   r+=(d-50)*0.05;
                   d=50;
          case 9:
          case 8:
          case 7:
          case 6:
          case 5:
          case 4:
                   r+=(d-30)*0.08;
                   d=20;
          case 3:
          case 2:  
                   r+=(d-10)*0.1;
                   d=10;            
          case 1:    
                   r+=(d-5)*0.15;
                   d=5;

          default:

                   r+=d*0.2;    

         }

        //输出结果

         printf("订单总额为%5.2f万元的毛利润是%5.2f万元\n",t,r);

         return 0;    
}