bdavis:
Ok, so...today being a very special day...took my first breathe on this very day exactly 2 score years ago...

In appreciation to being alive and thanking God for such a wonderful and marvelous journey so far, I have decided to drop $30 to a sharp nairalander...

But cos na me, the person go proof say hin sharp.

So here goes:


Decode this:

LZVA73AZQ5LTMVU


tell us what it is, the value associated with it, and get $30

NB: you have to tell us how you arrived at the value.

dhtml18:
@guru, you actually wrote something very nice this time. I however was not too pleased with the syntaxes, so i forked the code to - https://github.com/dhtml/jquerynigeriastatelga

Now this version is somewhat simpler for me to use, i cant say for others.

Nmeri17:
you wan gather your guys dem come obtain me?? Abeg! I work in the Lord's vineyard Awon permanent secretary Team no leave, no transfer. And my last post on this thread was meant to be funny; I'm sorry if you feel butthurt. Your Darling Andela does not rear weaklings.

dhtml18:

Ahem, Jamb Question!

Nmeri17:
^^^ Why won't you tire weakling! Andela rears weaklings!!

dhtml18:

True
@guru01, i have developed this library over a year ago. But i am just publishing it for public consumption for the first time - so should I call it version 5.0? because there has been many revisions since the first time I made it.
There is an a chat android application that is rather similar to whatsapp that i developed 2 years back, it runs on this platform. Nairaland developers that are very close to me have already seen this in action - and some of them already had the source code of the library even as far back as last year.
All i did not was to standardize the codes a bit and publish it. If you check the github link very well - https://github.com/dhtml/phpsocket.io - you will find out that it has been published on github like 6 months ago.

slysoft:
@jacob05

Impressive script, i just learned something.

The loop in your script started from 2 till 5001... can you explain why?

slysoft:



Hi @jacob05,

I designed the php script @slyrox uploaded for the problem... as @Lyphiard rightly mentioned that the problem @OP posted came from a competition in USA which we solved in our team before sharing it nairaland to develop other programmers on this platform.

The accepted solution/result on the competition portal was "5.08845978368001251890" which is the same as the result of our script/algorithm and different from the solution/result of your script/algorithm "5.088459783680012467179075225".

However, this is how we solved it or how we got the algorithm;

1
Looking at the equation posted;
x^2 + (5/3)x^3 + (23/12)x^4 + (119/60)x^5 ... + (2*[5000!-1]/[5000!])x^5000
it can easily be translated into a summation equation
which is to sum the result of ((2 * ([a+1]! – 1) / [a+1]! ) * x^(a+1) from a = 1 till 5000 where x is a constant
so for a = 1 we have x^2
a = 2 we have (5/3)x^3
sum the results till a = 5000

All we did was to translate the deduced summation equation into a script/algorithm. there was no reason behind using php, I could have used several other lang java, C#, VB, C, ruby, python

The result of the problem was also requested in the nearest 20 decimal places. why i added bcscale(25) is to control the number of trailing digits after the decimal and not affect the integrity of the 20th digit after the decimal. then i took off the trailing 5 digits after the 20th decimal

2: I will be willing to learn if you comeup with a much quicker or less-lame solution with regards to the posted problem. Thanks


BTW, My understanding of your script is that 35000 = 5000!; correct me if i am wrong.

from decimal import *
import math

x = Decimal('0.7893')

beginH = Decimal('1.0' )
beginB = Decimal('1.0' )

multi = Decimal('3' )
adder = Decimal('2' )

r = Decimal('0.0')
for i in range( 2, 5001 ) :
	r = r + ( ( beginH/beginB ) * Decimal( x ** i ) )
	
 	beginH = multi * beginH + adder
 	beginB = multi * beginB
 	multi +=1
 	adder +=1
print '{0:.21g}'.format(r)

olyjosh:


I begin to love dis python of a thing oo.
Your solution is quite brilliant n fast but your answer only converges upto d 15th decimal values. Unnecessary rounding in d system I guess.

dozies1:
Ok.
What is the principle behind making it go faster.

crotonite:
guyz, i finaly got it 100%

here is my code...

class Solution {
    public int solution(int[] A) {
        // write your code in Java SE 8
        if(A == null || A.length <= 0 || A.length > 100000)
        return -1;
        long r = 0;
        
        for(int i : A)
        r += i;
        
        long l = 0;
        
        for(int p = 0; p < A.length; p++){
            r -= A[p];
            if( r == l)
             return p;
             
            l += A[p] ;
            }
            return -1;
    }
}

thanks @febup for this challenge.

def solution(data ) :
	if not type(data ) is list or len(data ) <= 0: return -1
	lSum = 0; r = sum(data )
	for p in xrange(0, len( data ) ) :
		r -= data[p]
		if (r == lSum ) : return p
		lSum += data[p]
	return -1

dozies1:

here is my solution:

. 
 from math import factorial
 from decimal import Decimal

 n = 5000
 x = Decimal('0.7893')
 y = Decimal('0')

 for i in range(1, n+1):
     y = y + 2 * (x**i) * (factorial(i)-1) / factorial(i)
 print y
 

Please, would you mind telling me why mine is much more slower than yours.

Febup:


Tried you code works fine but does not allow for P == 0 or P == N-1 as mentioned in the instruction:
Sum of zero elements is assumed to be equal to 0. This can happen if P = 0 or if P = N−1.

mbatuku2:
Can someone recommend good php books for a beginner, especially written by someone in the academics? I'm open to any good author in the field that teach clearly too.

Febup:


I need someone to run your code for us. Over to you dhtml18 if you have some time to spare.

jacob05:

In Python

def solution(data):
lSum = 0; ret = []; rSum = sum(data )
if (rSum - data[0] == 0 ) : ret.append(1 )
lSum = data[0]
for p in xrange(1, len( data ) ) :
if (lSum == (rSum - (data[p] + lSum)) ) : ret.append(p)
lSum += data[p]
return ret
ar = [-1, 3, -4, 5, 1, -6, 2, 1 ]
result = solution(ar )
print result

Febup:
^^^
I don't have PHP setup on my machine so I tried your code on their site but it gave the error message below:

PHP Parse error: syntax error, unexpected '=' in func.php on line 6
Parse error: syntax error, unexpected '=' in func.php on line 6
RUNTIME ERROR (tested program terminated unexpectedly)

Detected some errors.

/* Your code*/
function solution($A) {
// write your code in PHP5.5
lSum = 0;
ret = [];
rSum = sum(A) )
if (rSum - A[0] == 0 ) : ret.append(1 )
lSum = A[0]
for p in xrange(1, len( A ) ) :
if (lSum == (rSum - (A[p] + lSum)) ) : ret.append(p)
lSum += A[p]
return ret
}

crotonite:
febup here is my solution.
i implemented the function solution(...) to return an int array of p's. plz check it out!

def solution(data):
  lSum = 0; ret = []; rSum = sum(data )
  if (rSum - data[0] == 0 ) : ret.append(1 )
  lSum = data[0]
  for p in xrange(1, len( data ) ) :
    if (lSum == (rSum - (data[p] + lSum)) ) : ret.append(p)
    lSum += data[p]
  return ret
  
ar = [-1, 3, -4, 5, 1, -6, 2, 1 ]
result = solution(ar )
print result

crotonite:
the prototype of my recent solution:

 import java.util.*;

public class Main
{
	public static void main(String[] args)
	{
		//int[] data = {-1, 3, -4, 5, 1, -6, 2, 1};

		int[] data = new int[100000];
		Random rand = new Random();
		
		 for(int i = 0; i  < data.length; i++){
			data[i] = rand.nextInt();
			//System.out.println(data[i]);
		}
		
		System.out.println("please wait...";
		double start = System.currentTimeMillis();
		int equill = solution(data, data.length);
		
		System.out.println();
		
		System.out.println("equillibrum at : " + equill);

			
		double stop = System.currentTimeMillis();

		System.out.println("it took " + ((stop - start)/1000) + "seconds to run "  );
	}





	private static int solution(int[] data, int n)
	{
		long lSum = 0, rSum = 0; //left and right summ of array
		int p = 0; //index of suspected equillibrum of array

		int ret = -1;
                
		//sum array 
		for (int i : data)
			rSum += i;
                
			
			//System.out.println("sum.." + rSum);
	

		//if first element is an aquillibrum, add to array
		if (rSum - data[p] == 0)
		{
			
		return p; }

		lSum = data[0];	
                 
		for (int i = 0; i < n - 1; i++)
		{
			p++;
			
			//System.out.println("rsum: " + rSum + " lsum: " + lSum);
			
			rSum -= (data[p] + lSum);

			if (lSum == rSum)
			{

				return p;

			}//endif

			//increment leftsum
			lSum += data[p];
		}
                
		
		return ret;
	}
} 

larisoft:



This runs in O(n).

public static int solution2(int[] nums, int n){

int sum = sum(nums);

int i = 0;
int left_sum = 0;

while(i < nums.length){

int right_sum = sum - left_sum - nums[i];

if(right_sum == left_sum){
return i;
}

left_sum+= nums[i];
i++;

}

return -1;

}

private static int sum(int[] arr){

int a = 0;
for(int i: arr){
a+=i;
}
return a;
}

eNelo:
Sweetie, you can do learn anything on your own if you're ready to sweat. Grit is the only requirement thats indispensable to learn any technology.

The first few months of programming are usually quite tricky, but the key is this: keep coding. Books are great yes, but you could read 10 great books and not be able to compete with a coder that has doled out 10 projects.

My advice to you would be this: identify something small that you need, and create the app. Let me explain: when I first started, I noticed how much I used the Windows Sticky Notes, but I always had to write the dates myself. So I used Java to create another version of sticky notes.
It can get really confusing: not really understanding what youre doing. But it gets better pretty fast.

Also, finishing one project is no excuse to sit back. As youre finishing one, start another. Making projects is the fastest and most thoroug route to coding well IMO.

Slyr0x:
^^I will check that in a bit.

Updated


Lol, Jacob05 , you've got the wrong answer. I ran your code, saw the first 5.088459******* and then assumed you got it right.

Slyr0x:
Great job Jacob05. Your solution is more efficient.

Solution below as solved by someone on my team

bcscale(25);
$itotal = "0";
$xt = "0.7893";$ii = 1;
$yyy = "1";while ($ii <= 5000)
{
    $y2 = intval(gmp_add($yyy,"1");
    $y3 = strval($y2);
    
    $factit = gmp_fact($y2);
    
    $at = bcdiv(gmp_mul("2", gmp_sub($factit,"1"), $factit);
    $xtpow = bcpow($xt,$y2);
    
    $anst = bcmul($at,$xtpow);
    
    $itotal = bcadd($itotal,$anst);
    $yyy = gmp_add($yyy,"1";
    $ii++;
    
}

echo "Solution: " .$itotal."<br />";

olyjosh:
Please guys, do submit your computation time and memory usage for these solutions. Anybody could have solve it, but let's see how effective is your solution in term of space and time.
You can log ur codes here https://ideone.com/ or on your machine(please do state the machine processor and memory configuration).
Thanks

Slyr0x:


Calm your tits.

I solved the challenges before posting here, and the answer for this is 5.08xxxxxxxxxxxxxxxxxx.

I will release the write up later .

Cheers Kiddo,

R.