我编写了一则c++程序，可是电脑运行时显示有错，我没看出有错，请高人指点

Question

#include "stdio.h"
int main()
{
int t,s;
float I,award;
printf("enter I=");
scanf("%f",&award);
t=100000;
s=I/t;
if(s>=10)
{ award=t*0.1+t*0.075+2*t*0.05+2*t*0.03+4*t*0.015+(I-10*t)*0.01;
printf("award=%f",award);
}
else
{
switch(s)
case 0: award=I*0.1;
printf("award=%f",award);break;
case 1: award=t*0.1+(I-t)*0.075;
printf("award=%f",award);break;
case 2:
case 3: award=t*0.1+t*0.075+(I-2*t)*0.05;
printf("award=%f",award);break;
case 4:
case 5: award=t*0.1+t*0.075+2*t*0.05+(I-4*t)*0.03;
printf("award=%f",award);break;
case 6:
case 7:
case 8:
case 9: award=t*0.1+t*0.075+2*t*0.05+2*t*0.03+(I-6*t)*0.015;
printf("award=%f",award);
}
return 0;
}
电脑运行结果上显示错误：
error C2043: illegal break
error C2046: illegal case。
这很明显llegal break指的是break有误，illegal case指的是case4/6/7/8有误，但这都是正确的，我一直不能理解电脑运行结果，望高人指出其中的错误，谢谢

Answer 1

你的问题就在于，你的case语句没有用大括号括起来，编译器没有分出来，他是switch的语句。
加一个大括号，就没错误了。
#include "stdio.h"
int main()
{
int t,s;
float I,award;
printf("enter I=");
scanf("%f",&award);
t=100000;
s=I/t;
if(s>=10)
{ award=t*0.1+t*0.075+2*t*0.05+2*t*0.03+4*t*0.015+(I-10*t)*0.01;
printf("award=%f",award);
}
else
{
switch(s)
{case 0: award=I*0.1;
printf("award=%f",award);break;
case 1: award=t*0.1+(I-t)*0.075;
printf("award=%f",award);break;
case 2:
case 3: award=t*0.1+t*0.075+(I-2*t)*0.05;
printf("award=%f",award);break;
case 4:
case 5: award=t*0.1+t*0.075+2*t*0.05+(I-4*t)*0.03;
printf("award=%f",award);break;
case 6:
case 7:
case 8:
case 9: award=t*0.1+t*0.075+2*t*0.05+2*t*0.03+(I-6*t)*0.015;
printf("award=%f",award);
}}
return 0;
}

Answer 2

我帮你修改了。不只是break和case问题，这两个是因为你的switch后面没有加大括号{}..
#include "stdio.h"
int main()
{
int t,s;
double I,award;
printf("enter I=");
scanf("%lf",&I);
t=100000;
s=(int)(I/(double)t);
if(s>=10)
{ award=t*0.1+t*0.075+2*t*0.05+2*t*0.03+4*t*0.015+(I-10*t)*0.01;
printf("award=%lf",award);
}
else
{
switch(s)
{
case 0:
award=I*0.1;
printf("award=%lf",award);
break;
case 1: award=t*0.1+(I-t)*0.075;
printf("award=%lf",award);break;
case 2:
case 3: award=t*0.1+t*0.075+(I-2*t)*0.05;
printf("award=%lf",award);break;
case 4:
case 5: award=t*0.1+t*0.075+2*t*0.05+(I-4*t)*0.03;
printf("award=%lf",award);break;
case 6:
case 7:
case 8:
case 9: award=t*0.1+t*0.075+2*t*0.05+2*t*0.03+(I-6*t)*0.015;
printf("award=%lf",award);
}
}
return 0;
}