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：按照题目中要求的数据分步计算，多次用到else if的判断

注意事项:等号的顺序

参考代码:

#include<stdio.h>

int main()

{

    int i;

    int bouns,bouns1,bouns2,bouns3,bouns4,bouns5;

    scanf("%d",&i);

    bouns1=100000*0.1;

    bouns2=bouns1+100000*0.075;

    bouns3=bouns2+200000*0.05;

    bouns4=bouns3+200000*0.03;

    bouns5=bouns4+400000*0.015;

    if(i<=100000){

        bouns=i*0.1;

        printf("%d",bouns);

    }

    else if (i<=200000){

        bouns=bouns1+(i-100000)*0.075;

        printf("%d",bouns);

    }

    else if (i<=400000){

        bouns=bouns2+(i-200000)*0.05;

        printf("%d",bouns);

    }

    else if (i<=600000){

        bouns=bouns3+(i-400000)*0.03;

        printf("%d",bouns);

    }

    else if (i<=1000000){

        bouns=bouns4+(i-600000)*0.015;

        printf("%d",bouns);

    }

    else if (i>1000000){

        bouns=bouns5+(i-1000000)*0.01;

        printf("%d",bouns);

    }

}