第二章课堂练习-1

#include<stdio.h>
int main(void)
{
   float n,m;
   printf("Input n:");
   scanf("%f",&n);
   if(n<2000)
   {
       m=500;
       printf("%f",m);
   }
   else if(n>=2000&&n<5000)
   {
       m=500+(n-2000)*0.08;
       printf("%f",m);
   }
   else if(n>=5000&&n<10000)
   {
       m=500+3000*0.08+(n-5000)*0.1;
       printf("%f",m);
   }
   else
   {
       m=500+3000*0.08+500+(n-10000)*0.12;
       printf("%f",m);
   }
   return 0;

}#include<stdio.h>
int main()
{
int a,i;
printf("Enter a:");
scanf("%d", &a);
i = a / 1000;
switch (i)
{
case 1:  a = 500;
printf("%d", a);
break;

case 2:  a = a + (a - 2000) * 0.8;
printf("%d", a);
break;
case 3:
case 4:

case 5: a = a + 240 + (a - 5000) * 0.1;
printf("%d", a);
break;
case 6:
case 7:
case 8:
case 9:
case 10:a = a + 240 + 500 + (a - 10000) * 0.12;
printf("%d", a);
break;
default:
a = a + 240 + 500 + (a - 10000) * 0.12;
printf("%d", a);
}
return 0;

}

计算工资。已知某公司员工的工资底薪为500元，员工销售的软件金额与提成方式如下：
销售额<2000                       没有提成
2000<=销售额<5000             提成8%