编写简单程序实现如下功能：有2个电阻并联，其阻值分别为10欧姆，30欧姆，求并联后的电阻。提示：并联电阻的公式是：r=1/(1/r1+1/r2).
谢谢各位的帮忙

float fun(int r1,int r2)
{
   float r;
   
   r = 1 / (1.0 / r1 + 1.0 / r2);
   return r;
}

#include<stdio.h>

double series(double r1,double r2);
double parallel(double r1,double r2);
int main()
{
  double r1,r2;
  int flag=0;
  printf("\nR1:");
  scanf("%lf",&r1);
  printf("\nR2:");
  scanf("%lf",&r2);
  printf("\n连接方式(0=series;1=parallel):");
  scanf("%d",&flag);
  if(flag==0)
     printf("%f\n",series(r1,r2));
  else
     printf("%f\n",parallel(r1,r2));
  return 0;
}
double series(double r1,double r2)
{
    double sum=r1+r2;
    return sum;
}

double parallel(double r1,double r2)
{
    double sum=1.0 / (1.0 / r1 + 1.0 / r2);
    return sum;
}

大家帮帮我啊  越简单越好的，第一个怎么不行啊

#include<stdio.h>
void main()
{
  float r1,r2,r;
  printf("Enter two Resistance:\n");
  scanf("%f,%f",&r1,&r2);          //输入两个电阻值   
  r=(r1*r2)/(r1+r2);               //计算公式
  printf("The Parallel Resistance is %f\n",r);

}

#include<stdio.h>
float fun(float x,float y)
{
float r;r=x*y/(x+y);
return r;
}
main()
{
float r,r1,r2;
scanf("%f,%f",&r1,&r2);
r=fun(r1,r2);
printf("%f",r);
}