Algorithms

Ian Uy wrote:

this will produce a relatively small number. And if I subtract that small number to value 2, the result will not be different. It is still 2.36 x 10^19,709. Right?

Value 2 = 2.36 x 10^19,709

Value 1 = 1.36 x 10^58

Ian Uy wrote:

Value 1 = 1.36 x 10^58
Value 2 = 2.36 x 10^19,709

Is there any mathematical technique or approach to simplify this?

Ian Uy wrote:

Is there any mathematical technique or approach to simplify this?

Roger Wright wrote:

You're still going to have a hell of a time getting a computer to calculate a number that larg

SatyaVas wrote:

......should generate sixteen combinations..

// VanillaConsole.cpp : Defines the entry point for the console application.
//
 
#include "stdafx.h"
#include <iostream>
#include <algorithm>
 
void main()
{
	using namespace std;
 
	char initial[] = "ABCD";
	char *pEnd = initial + sizeof(initial)/sizeof(initial[0])-1;
 
	do
	{
		cout << initial << endl;
	}
	while (next_permutation(initial, pEnd));
}

winburn wrote:

does anyone have a good formula for calculating azimuth between two city lat and lons

main()
{
	for(int i=0;i<n;i++)>
	{
		for(int j=0;j<n;j++)>
		{
			for(int k=0;k<n;k++)>
			{
				// do the something from func2
			}
			// do the something from func 1
		}
	}
}

main()
{
	int count = 0;
	for(int i=0;i<n;i++)>
	{
		for(int j=0;j<n;j++)>
		{
			for(int k=0;k<n;k++)>
			{
				count++;
			}
		}
	}
}

main()
{
	int count = 0;
	for(int i=0;i<n;i++)>
	{
		count++;		
	}

	for(int i=0;i<n;i++)>
	{
		count++;		
	}

	for(int i=0;i<n;i++)>
	{
		count++;		
	}
}

leppyr64 wrote:

When you nest loops you multiply the complexity.

main()
{
	List<int[]> n = new List<int[]>();
	
	for(int i=0;i<n.count;i++)>
	{
		n[i]=new int[100];
		for(int j=0;j<n[i].length;j++)>
		{
			for(int k=0;k<n[i].length;k++)>
			{
				
			}
		}
	}
}