One yuan per bottle of soda, two empty bottles can replace a bottle of soda, the existing 20 yuan, the maximum can drink how many bottles?

If this problem is not considered in C language programming, purely mathematical thinking, the problem is very simple. We can list a few, 2 yuan can drink 3 bottles, 3 yuan can drink 5 bottles, 4 yuan can drink 7 bottles, 5 Yuan can drink 9 bottles, we are not difficult to find if there is n yuan, you can drink 2*n-1 bottles. The problem is very simple.

#define N
int main ()
{
	printf ("Drink up to%d bottles \ n", 2*n-1);
	return 0;
}

This is a method that also has the logic to complete this question:

1 yuan per bottle of soda, two empty bottles for a bottle of soda. 2n-1 bottle (n yuan);
int main ()
{
	int money = 0;
	int count = 0;
	printf ("Please enter money:");
	scanf ("%d", &money);
	Count = money;
	while (1)
	{
		if (2!= 0)
		{
			count = count + money-1;
			break;
		else
		{Money
			= MONEY/2;
			Count + = money;
		}
	}
	printf ("Altogether can buy:%d\n", count);
	System ("pause");
	return 0;
}