Question

关于switch case的一个很简单的问题

编写一个程序
要求按工资少于1000，大于1000少于2000，大于2000区分
必须使用switch case语句
急

Answer 1

#include<stdio.h>
main()
{ int a,b,x=0,y=0,z=0;/*a是输入的工资,x是少于1000的人数,y是1000~2000的人数,z是>2000的人数*/
do
{ scanf("%d",&a);
switch(b=a/1000)
{case 0: ++x;break;
case 1: ++y;break;
default: ++z;}
scanf("%d",&a);
}while(a>0);
printf("x=%d y=%d z=%d",x,y,z);
}
本程序以输入数据<=0时作为结束条件
可以同时处理多个数据并记录

Answer 2

#include<stdio.h>
void main()
{
float i;
char c='y';
while(c=='y'||c=='Y')
{
clrscr();
printf("\n\nPlease input your salary:");
scanf("%f",&i);
if(i<0)
{
printf("The salary you input is not exist!");
goto l;
}
switch((int)(i/1000))
{
case 0:
printf("The salary is %.0f,it's smaller than 1000",i);goto l;
case 1:
printf("The salary is %.0f,it's bigger than 1000,but smaller than2000",i);goto l;
default:
printf("The salary is %.0f,it's bigger than 2000",i);goto l;
}
l:
printf("\nDo you want to continue?('y' or 'n')");
scanf("%s",&c);
}
printf("\n\t\t\tThanks for using!\n");
getch();
}

不知道能不能达到你的要求??

Answer 3

#include<stdio.h>
#include<conio.h>

int main()
{
int n;
printf("Input:");
scanf("%d",&n);
switch(n/1000)
{
case 0:
printf("less than 1000!\n");break;
case 1:
printf("more than 1000 and less than 2000!\n");break;
default:
printf("more than 2000!\n");break;
}
getch();
return 0;
}

Answer 4

#include<stdio.h>
void main()
{
int x;
printf("Input:");
scanf("%d",&x);
switch(x/1000)
{
case 0: ....;break; //x<1000,则x/1000<1;
case 1: ....;break; //1000<x<2000,则1<x/1000<2;
default:....;break; //x>2000 的情况
}
}

Answer 5

妙