Question

c语言！求助！要用函数和函数调用！

编程！要求让读者选“R”“C”
如果选R，问他用了多少度电(int)，然后算出他要交多少钱（double型），方程是pay=4.00+0.03*电量
如果选C，问他用了多少度电（int），如果小于1000度，pay=40,如果大于1000
pay=(电量-1000)*0.025+40.00，再告诉他要交多少钱（double型）。。。
要求两个选项都要用函数和函数调用， every function returns a "double" and takes an input argument of type "int".

Answer 1

#include <stdio.h>
double R(int r){
return 4.00+0.03*r;
}
double C(int c){
if(c<1000)
return 40.0;
else
return (c-1000)*0.025+40.00;
}
main(){
int t;

char ch;

printf("\nPls choice R/C: ");
ch=getchar();

printf("\n用了多少度电:");
scanf("%d",&t);

if(ch=='R'||ch=='r')
printf("要交%0.2lf元钱",R(t));

else
printf("要交%0.2lf元钱",C(t));
getch();
}

Answer 2

#include<stdio.h>
#include<conio.h>

double R()
{
int d;
do
{
printf("输入电量(整数):");
scanf("%d", &d);
}while(n<0);
return 4.00+0.03*d;
}

double C()
{
int d;
do
{
printf("输入电量(整数):");
scanf("%d", &d);
}while(n<0);
if(n<1000)
return 40.0;
else
return (n-1000)*0.025+40.0;
}

void main()
{
char ch;
double t;
do
{
printf("Please input R or C :");
ch = getch();
}while(ch != 'R' && ch != 'C');
if(ch == 'R') t = R();
else t=C();
printf("你的缴费金额：%.2f\n", t);
getch();
}

Answer 3

#include<stdio.h>
#include<conio.h>

double R(int d)
{
return 4.00+0.03*d;
}

double C(int d)
{
if(d<1000)
return 40.0;
else
return (d-1000)*0.025+40.0;
}

int main()
{
char c;
int dg;
double p;
do
{
printf("Plz input R or C :\n");
c=getch();
}
while(c!='R'&&c!='C'&&c!='r'&&c!='c');
printf("Plz input degree:\n");
scanf("%d",&dg);
if(c=='R'||c=='r')
p=R(dg);
else
p=C(dg);
printf("your fee is %.2f.\n", p);
return 0;
}