c语言实现计税函数 c语言算税
C语言编写个人所得税?
代码文本:
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#include "stdio.h"
int main(int argc,char *argv[]){
double x,tax;
printf("Please enter the number salary, negative end...\n");
while(scanf("%lf",x),x=0){
if(x=5000)
tax=(x-5000)*0.2+4200*.03;
else if(x=800 x5000)
tax=(x-800)*.03;
else
tax=0;
printf("You should pay %.2f yuan.\n",tax);
}
return 0;
}
C语言计算个人所得税 编程
#include stdio.h
#include stdlib.h
int jishu(double x)
{
if(0xx=500)
return 1;
else if(500xx=2000)
return 2;
else if(2000xx=5000)
return 3;
else if(5000xx=20000)
return 4;
else if(20000xx=40000)
return 5;
else if(40000xx=60000)
return 6;
else if(60000xx=80000)
return 7;
else if(80000xx=100000)
return 8;
else
return 9;
}
main()
{
double rate[10]={0.0,0.05,0.1,0.15,0.2,0.25,0.3,0.35,0.4,0.45};
int a[10]={0,0,25,125,375,1375,3375,6375,10375,15375};
double n,m,l;
int i;
printf("请输入工资:");
scanf("%lf",l);
if(l=3500)
printf("您不用交税\n");
else
{
n=l-3500.0;
i=jishu(n);
m=n*rate[i]-a[i];
printf("应缴个人所得税:%.2lf\n实发工资额:%.2lf\n",m,l-m);
}
}
这是按你说的计算方法
C语言,个人所得税计算,求大神回答- -悬赏有点少
#includestdio.h
double IIT(int money,int nation)
{
double iit = money;
switch(nation)
{
case 1 : iit = money - 1000 - 3500 ;break;
case 0 : iit = money - 1000 - 4800 ;break;
default: printf("输入有误!\n");
}
if(iit =1500) iit = iit*0.03;
else if(iit1500iit=4500) iit = (iit*0.1 - 105);
else if(iit4500iit=9000) iit = (iit*0.2 - 555);
else if(iit9000iit=35000) iit = (iit*0.25 -1005);
else if(iit35000iit=55000) iit= (iit*0.3 -2755);
else if(iit55000iit=80000) iit= (iit*0.35 -5505);
else iit = (iit*0.45 - 13505);
return iit;
}
int main()
{
int money,nation;
printf("请确定你的国籍: 1.中国 0.外籍\n");
scanf("%d",nation);
printf("请输入您的工资: ");
scanf("%d",money);
if(nation == 1){
if(money = 4500)
printf("您不需要缴纳个人所得税。\n");
else
printf("您要缴纳的个人所得税为: %.0f",IIT(money,nation));
}
if(nation == 0)
{
if(money = 5800)
printf("您不需要缴纳个人所得税。\n");
else
printf("您要缴纳的个人所得税为: %.0f",IIT(money,nation));
}
return 0;
}
c语言实现计税函数
#includestdio.h
float getTax(float salary)
{
float tax;
if(salary=10000)
tax=salary*0.1;
else if(salary10000 salary=20000)
tax=10000*0.1+(salary-10000)*0.12;
else if(salary20000 salary=30000)
tax=10000*0.1+10000*0.12+(salary-20000)*0.15;
else if(salary30000 salary=40000)
tax=10000*0.1+10000*0.12+10000*0.15+(salary-30000)*0.18;
else if(salary40000)
tax=10000*0.1+10000*0.12+10000*0.15+10000*0.18+(salary-40000)*0.2;
return tax;
}
int main()
{
float salary,tax;
scanf("%f",salary);
tax=getTax(salary);
printf("tax=%f\n",tax);
return 0;
}
编写c语言程序实现税费的计算。输入一个奖金数,求应交税款及实得奖金数
#includestdio.h
int main()
{
float a ,b,c;
printf("请输入奖金数:");
scanf("%f",a);
if (a 500)
printf("应缴税款:%f\n实得奖金数:%f",a*0/100,a*100/100);
else if(a=500a1000)
printf("应缴税款:%f\n实得奖金数:%f", a * 3/ 100, a * 97 / 100);
else if(a=1000a2000)
printf("应缴税款:%f\n实得奖金数:%f", a * 5 / 100, a * 95 / 100);
else if(a=2000a5000)
printf("应缴税款:%f\n实得奖金数:%f", a * 8 / 100, a * 92 / 100);
else
printf("应缴税款:%f\n实得奖金数:%f", a * 12 / 100, a * 88 / 100);
return 0;
}
关于纳税的C语言程序
其一:
j=(x-3500)*(3%);
(3%)C语言中不能识别3%之类的、可以改为0.03
然后:
如果if后面的语句在两句或两句以上要用{}括起来
else也是
本文标题:c语言实现计税函数 c语言算税
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