数学上相等的变换，计算机似乎不一定

本题要求编写程序，计算序列部分和 1 - 1/4 + 1/7 - 1/10 + ... 直到最后一项的绝对值不大于给定精度eps。

输入格式:

输入在一行中给出一个正实数eps。

输出格式:

在一行中按照“sum = S”的格式输出部分和的值S，精确到小数点后六位。题目保证计算结果不超过双精度范围。

The standard solution of  the book:

Marked by B:
int main(void)
{
	int denominator,flag;
	double esp,item,sum;
	scanf("%f",&esp);
	flag=1;
	denominator=1;
	sum=0;
	item=1.0;
	do
	{
		item=flag*1.0/denominator;
		sum=sum+item;
		flag=-flag;
		denominator=denominator+3;
	}while(fabs(item)>esp);
	printf("sum = %.6f\n",sum);
}

Something I wrote:

Marked by A :
int main()
{
  double eps;
  scanf("%lf",&eps);
  double sum=0;
  int i=0;
  while(1.0/(3*i+1)>=eps)
  {
  	sum+=pow(-1,i)*1.0/(3*i+1);
  	i++;
  }
  printf("sum = %.6lf",sum);
}

Two different series of codes, and you are smart enough to find the difference quickly—one uses the loop of "while" while another uses the loop of " do while". And this is because two different understanding concerning the title: Does the last item include the primary item?  

But the real pony here is that if you amend the former codes into the following one-C:

Marked by C:
int main()
{
  double eps;
  scanf("%lf",&eps);
  double sum=0;
  int i=0;
  do
  {
  	sum+=pow(-1,i)*1.0/(3*i+1);
  	i++;
  }while(1.0/(3*i+1)>=eps);
  printf("sum = %.6lf",sum);
  return 0；
}

C is not equal to B.

And something interesting is they both pass the PTA's test!

I hope somebody can tell me the difference that I don't know about them. Thank you!