#include<stdio.h>
#include<math.h>
#include<string.h>
main()
{
float jiangj;
long int lirun;
clrscr();
printf("Please intput your lirun:");
scanf("%ld",&lirun);
if(lirun<=100000)
jiangj=lirun*0.1;
else if(lirun>100000&&lirun<=200000)
      jiangj=100000*0.1+((lirun-100000)*0.075);
      else if(lirun>200000&&lirun<=400000)
            jiangj=(100000*0.1+((lirun-100000)*0.075))+((lirun-200000)*0.05);
            else if(lirun>400000&&lirun<=600000)
                  jiangj=((100000*0.1+((lirun-100000)*0.075))+((lirun-200000)*0.05))+(lirun-400000)*0.03;
                  else if(lirun>600000&&lirun<=1000000)
                        jiangj=(((100000*0.1+((lirun-100000)*0.075))+((lirun-200000)*0.05))+(lirun-400000)*0.03)+(1000000-600000)*0.015;
                        else if(lirun>1000000)
                              jiangj=((((100000*0.1+((lirun-100000)*0.075))+((lirun-200000)*0.05))+(lirun-400000)*0.03)+(1000000-600000)*0.015)+(lirun-1000000)*0.01;
       printf("%.2f",jiangj);

getch();
}

#include<stdio.h>
#include<math.h>
#include<string.h>
main()
{
double jiangj;
long int lirun;
printf("Please intput your lirun:");
scanf("%ld",&lirun);
if(lirun<=100000)
jiangj=lirun*0.1;
else if(lirun>100000&&lirun<=200000)
      jiangj=100000*0.1+((lirun-100000)*0.075);
      else if(lirun>200000&&lirun<=400000)
            jiangj=(100000*0.1+((lirun-100000)*0.075))+((lirun-200000)*0.05);
            else if(lirun>400000&&lirun<=600000)
                  jiangj=((100000*0.1+((lirun-100000)*0.075))+((lirun-200000)*0.05))+(lirun-400000)*0.03;
                  else if(lirun>600000&&lirun<=1000000)
                        jiangj=(((100000*0.1+((lirun-100000)*0.075))+((lirun-200000)*0.05))+(lirun-400000)*0.03)+(1000000-600000)*0.015;
                        else if(lirun>1000000)
                              jiangj=((((100000*0.1+((lirun-100000)*0.075))+((lirun-200000)*0.05))+(lirun-400000)*0.03)+(1000000-600000)*0.015)+(lirun-1000000)*0.01;
       printf("%.2f",jiangj);

getch();
}

看的眼花

clrscr在conio.h里面。
函数名: clrscr
功能: 清除文本模式窗口
用法: void clrscr(void);

若想用，加#include <conio.h>

恩，我试试，谢谢啊

这个源码太难看了！！！

#include<stdio.h>
#include<math.h>
#include<string.h>
main()
{
float jiangj;
long int lirun;

printf("Please intput your lirun:");
scanf("%ld",&lirun);
int lr;
lr=(lirun-1)/10000;
switch (lr){
case 0:jiangj=lirun*0.1;break;
case 1:jiangj=100000*0.1+((lirun-100000)*0.075);break;
case 2:
case 3:jiangj=(100000*0.1+((lirun-100000)*0.075))+((lirun-200000)*0.05);break;
case 4:
case 5:jiangj=((100000*0.1+((lirun-100000)*0.075))+((lirun-200000)*0.05))+(lirun-400000)*0.03;break;
case 6:
case 7:
case 8:
case 9:jiangj=(((100000*0.1+((lirun-100000)*0.075))+((lirun-200000)*0.05))+(lirun-400000)*0.03)+(1000000-600000)*0.015;break;
default:jiangj=((((100000*0.1+((lirun-100000)*0.075))+((lirun-200000)*0.05))+(lirun-400000)*0.03)+(1000000-600000)*0.015)+(lirun-1000000)*0.01;break;   
}      
printf("%.2f",jiangj);
getchar();
}

试试这么写结果对不对~

对的，谢谢啊，我刚学，不怎么会