平淡无奇的解法

解题思路:

注意事项:

参考代码:

#include <stdio.h>

#include <stdlib.h>

int main()

{

    double pro,bon;

    double bon1=1*0.1;//10w

    double bon2=1*0.1+1*0.075;//20w

    double bon3=1*0.1+1*0.075+2*0.05;//40w

    double bon4=1*0.1+1*0.075+2*0.05+2*0.03;//60w

    double bon5=1*0.1+1*0.075+2*0.05+2*0.03+4*0.015;//100w

    scanf("%lf",&pro);

    pro=pro/100000;

    if(pro<=1)

    {

        bon=pro*0.1;

    }

    else if(pro>1&&pro<=2)

    {

        bon=(pro-1)*0.075+bon1;

    }

    else if(pro>2&&pro<=4)

    {

        bon=(pro-2)*0.05+bon2;

    }

    else if(pro>4&&pro<=6)

    {

        bon=(pro-4)*0.03+bon3;

    }

    else if(pro>6&&pro<=10)

    {

        bon=(pro-6)*0.015+bon4;

    }

    else if(pro>10)

    {

        bon=(pro-10)*0.01+bon5;

    }

    bon=bon*100000;

    printf("%.0lf",bon);

    return 0;

}