4-11.c

来自「C语言程序设计及应用习题解析与上机指导 冶金工业出版社 谢乐军」· C语言 代码 · 共 26 行

main()
{
long  profit;
     int  grade;
     float  salary=500;
     printf("Input  profit: ");
     scanf("%ld", &profit);
     grade= (profit – 1) / 1000; 
/*将利润-1、再整除1000，转化成  switch语句中的case标号*/
switch(grade)
{ 
case  0:  break;                                /*profit≤1000 */
   case  1: salary += profit*0.1; break;     /*1000＜profit≤2000 */
   case  2: 
   case  3: 
   case  4: salary += profit*0.15; break;   /*2000＜profit≤5000 */
   case  5: 
   case  6: 
   case  7: 
   case  8: 
   case  9: salary += profit*0.2; break;     /*5000＜profit≤10000 */
   default: salary += profit*0.25;                 /*10000＜profit */
}
printf("salary=%.2f\n", salary);
}