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Answers for "algorithm converter in latex"

Whatever
0

using latex to write algorithm

\begin{program}
\mbox{A fast exponentiation procedure:}
\BEGIN \\ %
  \FOR i:=1 \TO 10 \STEP 1 \DO
     |expt|(2,i); \\ |newline|() \OD %
\rcomment{This text will be set flush to the right margin}
\WHERE
\PROC |expt|(x,n) \BODY
          z:=1;
          \DO \IF n=0 \THEN \EXIT \FI;
             \DO \IF |odd|(n) \THEN \EXIT \FI;
\COMMENT{This is a comment statement};
                n:=n/2; x:=x*x \OD;
             \{ n>0 \};
             n:=n-1; z:=z*x \OD;
          |print|(z) \ENDPROC
\END
\end{program}
Posted by: Guest on July-23-2020
0

algorithm converter in latex

using System;

private static void Main()
{
	int n;
	float x;
	float x1;
	float x2;
	float x3;
	float f;
	Console.Clear();
	f = 0F;
	cprintf("\nEnter number of Tries : ");
	cscanf("%d", n);
	while (n > 0)
	{
	 cprintf("\nValue for X  : ");
	 cscanf("%f", x);
	 cprintf("\nValue for X1 : ");
	 cscanf("%f", x1);
	 cprintf("\nValue for X2 : ");
	 cscanf("%f", x2);
	 cprintf("\nValue for X3 : ");
	 cscanf("%f", x3);

	 if (x1 <= x && x <= x2)
	 {
		f = ((x - x1) / (x2 - x1)) * (x3 - x1);
	 }
	 else
	 {
		if (x2 < x && x <= x3)
		{
			f = ((x - x2) / (x3 - x2)) * (x2 - x1);
		}
		else
		{
			if (x > x3)
			{
				f = x3;
			}
		}
	 }
	cprintf("\nResult is : %f\n",f);
	n--;
	}
	Console.ReadKey(true);
}
Posted by: Guest on September-30-2021
0

algorithm converter in latex

def binary_search(nums, key):
    return binary_search_helper(nums, key, 0, len(nums))
    
def binary_search_helper(nums, key, start_idx, end_idx):
    middle_idx = (start_idx + end_idx) // 2
    if start_idx == end_idx:
        return None
    if nums[middle_idx] > key:
        return binary_search_helper(nums, key, start_idx, middle_idx)
    elif nums[middle_idx] < key:
        return binary_search_helper(nums, key, middle_idx + 1, end_idx)
    else:
        return middle_idx
Posted by: Guest on September-16-2021

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