第二问:举例来说。price=10500,超10000的的按0.05的税来收,之后得是不超过10000且大于5000的部分按一定水率来收,以此类推
你看我的理解对不。如果有问题,HI我。/*表达的有点不清楚,如果x是20000,按10%算还是按20%算*/
#include<stdio.h>
int main(void)
{
double tax=0,money,m
int c
scanf("%lf",&money)
m=money
if(money/10000>8)
c=8
else
c=(int)money/10000
switch(c)//找到一个入口,顺次相加各个级应纳税额。
{
case 8:tax+=(money-80000)*0.35money=80000
case 7:
case 6:
case 5:
case 4:tax+=(money-40000)*0.30money=40000
case 3:
case 2:tax+=(money-20000)*0.20money=20000
case 1:tax+=(money-10000)*0.10money=10000
case 0:tax+=money*0.05break
default:printf("Data Error!\n")
}
printf("应纳税额:%.2f\n",tax)
printf("最终所得:%.2f\n",m-tax)
return 0
}
为了便于你验证程序执行结果:下面的可以多次执行,直到你输入的money不大于0.
#include<stdio.h>
int main(void)
{
while(1)
{
double tax=0,money,m
int c
printf("请输入全年应纳所得额数目:\n")
scanf("%lf",&money)
if(money<=0)
break
m=money
if(money/10000>8)
c=8
else
c=(int)money/10000
switch(c)//找到一个入口,顺次相加各个级应纳税额。
{
case 8:tax+=(money-80000)*0.35money=80000
case 7:
case 6:
case 5:
case 4:tax+=(money-40000)*0.30money=40000
case 3:
case 2:tax+=(money-20000)*0.20money=20000
case 1:tax+=(money-10000)*0.10money=10000
case 0:tax+=money*0.05break
default:printf("Data Error!\n")
}
printf("应纳税额:%.2f\n",tax)
printf("最终所得:%.2f\n",m-tax)
}
return 0
}
编程1void main()
{
int n
printf("请输入一个整数:")
scanf("%d",&n)
if(n%3==0&&n%5==0&&n%7==0)
printf("\n此数能被3,5,7整除\n")
else
printf("\n此数不能被3,5,7整除\n")
}
编程2#include "stdio.h"
void main()
{
long income
double fee=0.
printf("input income:")
scanf("%ld",&income)
if(income>2000)
{
fee+=(income-2000)*.2
income=2000
}
switch((income/400))
{
case 5:
case 4:
case 3:
fee+=(income-1200)*.08
income=1200
case 2:
fee+=(income-800)*.05
default:
fee+=0.
}
printf("tax fee is %.2lf.\n",fee)}编程4#include <stdio.h>
#include <stdlib.h>
void main() {
int n, i
printf("请输入n:")
scanf("%d", &n)
int sum = 0
for(i=1i<=ni++) sum += i
printf("和是: %d\n", sum)
}
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