一个用C语言解决的计算奖金的问题

我这法儿也行，就是麻烦点：

# include <stdio.h>

void main()

{

float I,P

printf("input li run:")

scanf("%f",&I)

if(I>=0 &&I<=100000)P=I*0.1

else if(I>100000&&I<=200000){P=100000*0.1+(I-100000)*0.075printf(":liyun%.2f\n",P)}

else if(I>200000&&I<=400000){P=100000*0.1+100000*0.075+(I-200000)*0.05printf(":liyun%.2f\n",P)}

else if(I>400000&&I<=600000){P=100000*0.1+100000*0.075+200000*0.05+(I-400000)*0.03printf(":liyun%.2f\n",P)}

else if(I>600000&&I<=1000000){P=100000*0.1+100000*0.075+200000*0.05+200000*0.03+(I- 600000)*0.015printf(":liyun%.2f\n",P)}

else if(I>1000000){P=100000*0.1+100000*0.075+200000*0.05+200000*0.03+400000*0.015+(I- 1000000)*0.01printf(":liyun%.2f\n",P)}

else printf("error!\n")

}

#include<stdio.h>

#define MIN(a,b) ((a)<(b)?(a):(b))

int main()

{float lr,jj

scanf("%f",&lr)

jj=0.1*MIN(lr,100000)

if(lr>100000)jj+=0.075*MIN(lr-100000,100000)

if(lr>200000)jj+=0.05*MIN(lr-200000,200000)

if(lr>400000)jj+=0.03*MIN(lr-400000,200000)

if(lr>600000)jj+=0.015*MIN(lr-600000,400000)

if(lr>1000000)jj+=0.05*(lr-1000000)

printf("%g\n",jj)

return 0

}

#include<stdio.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>=500&&a<1000)

printf("应缴税款：%f\n实得奖金数：%f", a * 3/ 100, a * 97 / 100)

else if(a>=1000&&a<2000)

printf("应缴税款：%f\n实得奖金数：%f", a * 5 / 100, a * 95 / 100)

else if(a>=2000&&a<5000)

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

}

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