输入999999和5000.02无法输出，求助。？

#include
//定义main函数
#define EPSILON 0.01
int main()
{
//请在此添加‘奖金发放’的代码,预期格式如下，可改动。
/*****************Begin******************/
float d,e,h,a,g,x,y,z;
double c;
int b;
scanf("%f",&a);
if(a==100000)
b=1;

if(100000<a && a<200000)
b=2;

if(a==200000)
b=3;    

if(200000<a && a<400000)
b=4;

if(a==400000)
b=5;

if(600000.00>a && a>400000.00)
b=6;

if(a==600000)
b=7;

if(600000<a && a<1000000)
b=8;

if(a==1000000)
b=9;

if(a>1000000)
b=10;
switch (b)
{
    case 1:

           c=a*0.1;
           printf ("%.2f",c);
           break;

    case 2:

           d=a-100000;
           e=d*0.075;
           h=100000*0.1;
           c=e+h;
           printf ("%.2f",c);
           break;

    case 3:

           d=a-100000;
           e=d*0.075;
           h=100000*0.1;
           c=e+h;
           printf ("%.2f",c);
           break;

    case 4:

           d=a-200000;
           e=d*0.05;
           h=(200000-100000)*0.075;
           g=100000*0.1;
           c=e+h+g;
           printf ("%.2f",c);
           break;

    case 5:

          d=a-200000;
           e=d*0.05;
           h=(200000-100000)*0.075;
           g=100000*0.1;
           c=e+h+g;
           printf ("%.2f",c);
           break;

    case 6:

           d=a-400000;
           e=d*0.03;
           h=(400000-200000)*0.05;
           g=(200000-100000)*0.075;
           x=100000*0.1;
           c=e+h+g+x;
           printf ("%.2f",c);
           break;

    case 7:
           d=a-400000;
           e=d*0.03;
           h=(400000-200000)*0.05;
           g=(200000-100000)*0.075;
           x=100000*0.1;
           c=e+h+g+x;
           printf ("%.2f",c);
           break;

    case 8:
           d=(a-600000)*0.015;
           e=600000-400000;
           h=e*0.03;
           g=(400000-200000)*0.05;
           x=(200000-100000)*0.075;
           y=100000*0.1;
           c=d+h+g+x+y;
           printf ("%1.2f",c);
           break;

    case 9:
           d=(a-600000)*0.015;
           e=600000-400000;
           h=e*0.03;
           g=(400000-200000)*0.05;
           x=(200000-100000)*0.075;
           y=100000*0.1;
           c=d+h+g+x+y;
           printf ("%g",c);
           break;



    case 10:

           d=(a-1000000)*0.01;
           z=(1000000-600000)*0.015;
           e=600000-400000;
           h=e*0.03;
           g=(400000-200000)*0.05;
           x=(200000-100000)*0.075;
           y=100000*0.1;
           c=d+z+h+g+x+y;
           printf ("%.2f",c);
           break;

}

/***************** End ******************/
return 0;

}

你没有小于10000的选项啊！如下即有结果了！

#include <stdio.h>
//定义main函数
#define EPSILON 0.01
int main()
{
    //请在此添加‘奖金发放’的代码,预期格式如下，可改动。
    /*****************Begin******************/
    float d,e,h,a,g,x,y,z;
    double c;
    int b;
    scanf("%f",&a);
    if(a < 100000)
    b=0;
    if(a==100000)
    b=1;

    if(100000<a && a<200000)
    b=2;

    if(a==200000)
    b=3;    

    if(200000<a && a<400000)
    b=4;

    if(a==400000)
    b=5;

    if(600000.00>a && a>400000.00)
    b=6;

    if(a==600000)
    b=7;

    if(600000<a && a<1000000)
    b=8;

    if(a==1000000)
    b=9;

    if(a>1000000)
    b=10;
    switch (b)
    {
        case 0:
            printf("没有a < 100000 的选项！\n" );
            break;
        case 1:

               c=a*0.1;
               printf ("%.2f",c);
               break;

        case 2:

               d=a-100000;
               e=d*0.075;
               h=100000*0.1;
               c=e+h;
               printf ("%.2f",c);
               break;

        case 3:

               d=a-100000;
               e=d*0.075;
               h=100000*0.1;
               c=e+h;
               printf ("%.2f",c);
               break;

        case 4:

               d=a-200000;
               e=d*0.05;
               h=(200000-100000)*0.075;
               g=100000*0.1;
               c=e+h+g;
               printf ("%.2f",c);
               break;

        case 5:

              d=a-200000;
               e=d*0.05;
               h=(200000-100000)*0.075;
               g=100000*0.1;
               c=e+h+g;
               printf ("%.2f",c);
               break;

        case 6:

               d=a-400000;
               e=d*0.03;
               h=(400000-200000)*0.05;
               g=(200000-100000)*0.075;
               x=100000*0.1;
               c=e+h+g+x;
               printf ("%.2f",c);
               break;

        case 7:
               d=a-400000;
               e=d*0.03;
               h=(400000-200000)*0.05;
               g=(200000-100000)*0.075;
               x=100000*0.1;
               c=e+h+g+x;
               printf ("%.2f",c);
               break;

        case 8:
               d=(a-600000)*0.015;
               e=600000-400000;
               h=e*0.03;
               g=(400000-200000)*0.05;
               x=(200000-100000)*0.075;
               y=100000*0.1;
               c=d+h+g+x+y;
               printf ("%1.2f",c);
               break;

        case 9:
               d=(a-600000)*0.015;
               e=600000-400000;
               h=e*0.03;
               g=(400000-200000)*0.05;
               x=(200000-100000)*0.075;
               y=100000*0.1;
               c=d+h+g+x+y;
               printf ("%g",c);
               break;



        case 10:

               d=(a-1000000)*0.01;
               z=(1000000-600000)*0.015;
               e=600000-400000;
               h=e*0.03;
               g=(400000-200000)*0.05;
               x=(200000-100000)*0.075;
               y=100000*0.1;
               c=d+z+h+g+x+y;
               printf ("%.2f",c);
               break;

    }

    /***************** End ******************/
return 0;
}