10题的第二问，救救孩子吧

将利润除以10万，对得到的整数进行switch处理

#include <stdio.h>
int main()
{
      int i,k;
      double f = 0;
      scanf("%d",&i);
      k =  i/100000;
      switch(k)
      {
      default:
          f+=(i-1000000)*0.01;
          i=1000000;
      case 7:
      case 8:
      case 9:
          f+=(i-600000)*0.015;
          i=600000;
      case 4:
      case 5:
      case 6:
          f+=(i-400000)*0.03;
          i=400000;
      case 2:
      case 3:
          f+=(i-200000)*0.05;
          i=200000;
      case 1:
          f+=(i-100000)*0.075;
          i=100000;
      case 0:
          f+=i*0.1;
      }
      printf("%.2lf",f);
    return 0;
}

参考：

#include "stdio.h"
int main()
{
    long int i;
    float bonus1,bonus2,bonus4,bonus6,bonus10,bonus;
    scanf("%ld",&i);
    bonus1=100000*0.1;
    bonus2=bonus1+100000*0.75;
    bonus4=bonus2+200000*0.5;
    bonus6=bonus4+200000*0.3;
    bonus10=bonus6+400000*0.15;
    if(i<=100000)
        bonus=i*0.1;
    else if(i<=200000)
        bonus=bonus1+(i-100000)*0.075;
    else if(i<=400000)
        bonus=bonus2+(i-200000)*0.05;
    else if(i<=600000)
        bonus=bonus4+(i-400000)*0.03;
    else if(i<=1000000)
        bonus=bonus6+(i-600000)*0.015;
    else
        bonus=bonus10+(i-1000000)*0.01;
    printf("bonus=%.2f",bonus);
}

用I除以100000后向上取整，用这个数去进行switch计算。
代码如下：

#include <stdio.h>
int main()
{
    int i,t;
    float y;
    scanf("%d",&i);
    t = ceil(i/100000.0); //向上取整
    switch(t)
    {
    case 1:
        y = i * 0.1; break;
    case 2:
        y = 100000*0.1 + (i-100000)*0.075; break;
    case 3:
    case 4:
        y = 100000*0.1 + 100000*0.075 + (i-200000)*0.05;
        break;
    case 5:
    case 6:
        y = 100000*0.1 + 100000*0.075 + 200000*0.05 + (i-400000)*0.03;
        break;
    case 7:
    case 8:
    case 9:
    case 10:
        y = 100000*0.1 + 100000*0.075 + 200000*0.05 + 200000*0.03 + (i-600000)*0.015;
        break;
    default:
        y = 100000*0.1 + 100000*0.075 + 200000*0.05 + 200000*0.03 + 400000*0.015 + (i-1000000)*0.01;
        break;
    }
    printf("%f",y);
    return 0;
}