因为常量值后面的若干语句中没有break,则接着运行下一个常量值(case 3)后面的若干语句,直到遇到break为止
#include<stdio.h>
int main(){
int a,t;
scanf("%d",&a);
switch(a/100000){
case 0:
t=a*0.1;
break;
case 1:
t=100000*0.1+(a-100000)*0.075;
break;
case 2:
case 3:
t=100000*0.1+100000*0.075+(a-200000)*0.05;
break;
case 4:
case 5:
t=100000*0.1+100000*0.075+200000*0.05+(a-400000)*0.03;
case 6:
case 7:
case 8:
case 9:
t=100000*0.1+100000*0.075+200000*0.05+200000*0.03+(a-600000)*0.015;
default:
t=100000*0.1+100000*0.075+200000*0.05+200000*0.03+400000*0.015+(a-1000000)*0.01;
}
printf("%d",t);
return 0;
}
ps:当a/100000=2时,因为常量值后面的若干语句中没有break,则接着运行下一个常量值(case 3)后面的若干语句,直到遇到break为止
同理,后面的4,6,7,8都是一样的
程序结束return 0;不可省。