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c語言求利潤

發布時間: 2023-03-07 08:02:19

Ⅰ C語言程序設計:獎金利潤提成

if分支或switch都可以實現,這個用switch比較適合一些。

Ⅱ C語言編程,利潤提成

(1)if語句
#include "stdio.h"

void main()
{
int i,j;
float a1,a2,a4,a6,a10,k;
a1=100000*0.1;
a2=a1+100000*0.075;
a4=a2+200000*0.05;
a6=a4+200000*0.03;
a10=a6+400000*0.015;
printf("請輸入利潤:");
scanf("%d",&i);
if(i<=100000)
k=i*0.1;
else if(i<=200000)
k=a1+(i-100000)*0.075;
else if(i<=400000)
k=a2+(i-200000)*0.05;
else if(i<=600000)
k=a4+(i-400000)*0.03;
else if(i<=1000000)
k=a6+(i-600000)*0.015;
else
k=a10+(i-1000000)*0.01;
printf("獎金是:%.2f",k);
}
(2)switch語句
#include "stdio.h"

void main()
{
int i,j;
float a1,a2,a4,a6,a10,k;
a1=100000*0.1;
a2=a1+100000*0.075;
a4=a2+200000*0.05;
a6=a4+200000*0.03;
a10=a6+400000*0.015;
printf("請輸入利潤:");
scanf("%d",&i);
j=i/100000;
switch(j){
case 0:
k=i*0.1;
break;
case 1:
k=a1+(i-100000)*0.75;
break;
case 2:
case 3:
k=a2+(i-200000)*0.05;
break;
case 4:
case 5:
k=a4+(i-400000)*0.03;
break;
case 6:
case 7:
case 8:
case 9:
k=a6+(i-600000)*0.015;
break;
default:
k=a10+(i-1000000)*0.01;
}
printf("獎金是:%.2f",k);
}

Ⅲ C語言 「企業發放的獎金根據利潤提成」 的問題

你scanf("&d",&i); 改為scanf("%f",&i); 就行,最後面的輸出用%g好一些,可以出去多餘的0

#include<stdio.h>
int main()
{
float i,J,a,b,c,d,e;
printf("請輸入利潤:I=");
scanf("%f",&i);
a=10000;
b=7500;
c=10000;
d=6000;
e=6000;
if(i<=100000) J=i*0.1;
else if(i<=200000) J=a+(i-100000)*0.075;
else if(i<=400000) J=a+b+(i-200000)*0.05;
else if(i<=600000) J=a+b+c+(i-400000)*0.03;
else if(i<=1000000) J=a+b+c+d+(i-600000)*0.015;
else
J=a+b+c+d+e+(i-1000000)*0.01;
printf("獎金總數:%g\n",J);
return 0;
}

Ⅳ C語言編程題 利潤提成

#include <stdio.h>
int main()
{
int z,i;
float sum=0;
printf("input money:");
scanf("%d",&i);
if(i<=100000)
z=1;
else if(i<=200000&&i>100000)
z=2;
else if(i<=400000&&i>200000)
z=4;
else if(i<=600000&&i>400000)
z=6;
else if(i<=1000000&i>600000)
z=10;
else if(i>1000000)
z=11;
switch(z)
{
case 11:sum+=(i-1000000)*0.01;i=1000000;
case 10:sum+=(i-600000)*0.015;i=600000;
case 6:sum+=(i-400000)*0.03;i=400000;
case 4:sum+=(i-200000)*0.05;i=200000;
case 2:sum+=(i-100000)*0.075;i=100000;
case 1:sum+=i*0.1;break;
}
printf("Bonus=%lf",sum);
return 0;
}
調好了 試試

Ⅳ C語言求銀行本息和利潤和

#include<stdio.h>
void main(){
double num,rate=0.00417,sum=0;
int i;
scanf("%lf",&num);
while(num<=0){
printf("輸入的存款金額必須大於0!\n");
scanf("%lf",&num);
}
for(i=1;i<=5;i++){
sum=((100+sum)*(1+rate));
}
printf("%.3f\n",sum);
}