C语言程序设计教程（第三版）课后习题5.8 （C语言代码）

解题思路:

注意事项:

参考代码:#include <stdio.h>

int main(){

int a,t;

scanf("%d",&a);

switch(a/100000)

{

case 0;

t=a*0.1;

break;

case 1:

t=100000*0.1+(a-100000)*0.075;

break;

case2:

case3:

t=100000*0.1+100000*0.075+(a-200000)*0.05;

break;

case 4:

case 5:

t=100000*0.1+100000*0.075+200000*0.05+(a-400000)*0.03;

case 6:

case 7:

case 8:

case 9:

t=100000*0.1+100000*0.075+200000*0.05+200000*0.03+(a-600000)*0.015;

default:

t=100000*0.1+100000*0.075+200000*0.05+200000*0.03+400000*0.015+(a-1000000)*0.01;

}

printf("%d",t);

return 0;

}