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 i,bouns;

    scanf("%d",&i);

    switch(i/100000)

    {

        case 0 : bouns = i*0.1 ; break; 

        case 1 : bouns = 100000*0.1 + (i-100000)*0.075; break;

        case 2 : bouns = 100000*0.1 + 100000*0.075 + (i-200000)*0.05;break;

        case 3 : bouns = 100000*0.1 + 100000*0.075 + (i-200000)*0.05;break;

        case 4 : bouns = 100000*0.1 + 100000*0.075 + 200000*0.05 + (i-400000)*0.03; break ;

        case 5 : bouns = 100000*0.1 + 100000*0.075 + 200000*0.05 + (i-400000)*0.03; break ;

        case 6 : bouns = 100000*0.1 + 100000*0.075 + 200000*0.05 + 200000*0.03 + (i-600000)*0.015; break ;

        default :bouns = 100000*0.1 + 100000*0.075 + 200000*0.05 + 200000*0.03 + 400000*0.015 + (i-1000000)*0.01;

    }

    printf("%d",bouns);

    return 0;

}