₦airaland Forum

kudaisi:
The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

Let us list the factors of the first seven triangle numbers:

1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors.

What is the value of the first triangle number to have over five hundred divisors?

Editing ...

kudaisi:
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.

/**
 *
 * @author olyjosh
 */
public class SumOfPrime {
    public static void main(String[] args) {
        int max = 2000000;

        //I implemented Sieve by taking true to be false n vice-versa, Just to avoid initial looping/flagging bla-bla
        boolean[] isPrime = new boolean[max];
        for (int i = 2; i*i < max; i++) {
            if (!isPrime[i]) {
                for (int j = i; i*j < max; j++) {
                    isPrime[i*j] = true;
                }
            }
        }

        long sum = 0;
        for (int i = 2; i < max; i++) {
            if (!isPrime[i]) sum+=i;
        }
        System.out.println("Sum of prime <" + max + " is " + sum);
    }
}

Sum of prime <2000000 is 142913828922

author=olyjosh post=33427863]

/**
 *
 * @author olyjosh
 */
public class PythagoreanTriplet {
    
    public static void main(String[] args) {
        int lim=1000/3;
        for (int i = 2; i <=lim; i++) {
            for (int j = i+1; j <= lim; j++) {
                int c = j*j+i*i,
                        b=j*j-i*i,
                        a=2*i*j;
                if(1000==a+b+c)System.out.println("product: "+(a*b*c));
            }
        }
    }
   
}

product: 31875000[/quote]Quite effective, check it out here http://ideone.com/Y3B3wP

Runtime: 0seconds
Memory; 0kb

kudaisi:
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,

(see image for inserts)
For example, (see image for inserts)

There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.

/**
 *
 * @author olyjosh
 */
public class PythagoreanTriplet {
    
    public static void main(String[] args) {
        int lim=1000/3;
        for (int i = 2; i <=lim; i++) {
            for (int j = i+1; j <= lim; j++) {
                int c = j*j+i*i,
                        b=j*j-i*i,
                        a=2*i*j;
                if(1000==a+b+c)System.out.println("product: "+(a*b*c));
            }
        }
    }
   
}

product: 31875000

blueyedgeek:

 
(function ()  {
var sum = 0;
for (var i = 0; i <= 100; i += 1) {
sum += i;
} 
return Math.pow(sum, 2);
} ()) -  (function ()  {
var sum = 0;
for (var i = 0; i <= 100; i += 1)  {
sum += Math.pow(i, 2);
} 
return sum;
} () ) 
// 25164150

Cool implementation bro. But the best approach to this isn't greedy approach. The sum arithmetic progression is what you can use in both case. These are formula you can easily derive using mathematical inductions.

Sn = {n(1+n)}/2
Sum of squres in AP is = {n(n+1)(2*n+1)}/6
where one 1 can alway be the first term if your series does not start from 1 and n is the last term

Sorry I wish I could delete the above implementation, It's buggy, BRB when I fix it

kudaisi:
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.

What is the 10 001st prime number?

public class _10001stPrime {
    
    static boolean isPrime(long n)
    {
        if ( n % 2 == 0 )return false;
        long sqrt = (long) Math.sqrt(n);
        sqrt= sqrt%2 == 0 ? sqrt-1 : sqrt; 
        for ( int i = 3; i < sqrt; i += 2 ) 
            if ( n%i == 0 )return false;
        return true;
    }
    
    public static void main(String[] args) {
        int i =1;
        int n =3;
        while(i<10001){
            if(isPrime(n))i++;
            n+=2;
        }
        System.out.println("10001st prime number: "+n);
    }
}

10001st prime number: 103575

I out for now - Gat exams to write tommorow. BRB After.

kudaisi:
The sum of the squares of the first ten natural numbers is,

12 + 22 + ... + 102 = 385
The square of the sum of the first ten natural numbers is,

(1 + 2 + ... + 10)2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

/**
 *
 * @author olyjosh
 */
public class DifferenceSumsSquare {
    
    public static void main(String[] args) {
        final int n=100;
        int sqOfSum = (n*(n+1)*(2*n+1))/6;
        int sumOfsq = (n*(1+n))/2; sumOfsq*=sumOfsq;
        System.out.println("Difference is "+(sumOfsq-sqOfSum));
    }
}

Difference is 25164150

Jregz:
in php
function getSum($u){
for ($uu = 0 ; $uu < 1000 ; $uu++){
if ($uu % 3 === 0 || $uu % 5 === 0 ){
$u += $uu ;
}
}
return $u
}
echo getSum(0) ; // 233168

Welcome to the show

@WhiZTiM check this out
Runs for 10seconds on 4gb RAM, 1.65GHz Duo Core processor

public class Primes
{
    static boolean isPrime(long n)
    {
        if ( n % 2 == 0 )return false;
        long sqrt = (long) Math.sqrt(n);
        sqrt= sqrt%2 == 0 ? sqrt-1 : sqrt; 
        for ( int i = 3; i < sqrt; i += 2 ) 
        {
            if ( n % i == 0 )return false;
        }
        return true;
    }
    
    public static void main(String[] args)
{

    long t = 600851475143L;
    long d = 2;
    while (1==1)
    {
        long tmp = 600851475143L / d;
        if ( t % tmp == 0 &&  isPrime(tmp) )
        {
            System.out.println("= " + tmp);
            break;
        }
        d++;
    }
}
}

WhiZTiM:
Comparing the Runtime of olyjosh's simple and elegant solution, (Java) with mine, (Python)
You can see some abominable things happening... Python 3x faster than Java. ....HOw?
Memorization. ....

http://ideone.com/XNVT5m ....Java (greedy) ...0.07seconds 320MB
http://ideone.com/VSTUQe .....Python (Dynamic Programming) ...0.02seconds 8MB

Java will regain its glory if DP/Memorization is used.

Cool. I will tryimplementing this with DP. Haaaa, but I doubt if i understand most of this python Types and sytax correctly.

WhiZTiM:
@olyjosh.
I am really impressed! ...Would love to chat up with you Sir.

Nice meeting you Bro

WhiZTiM:
. . . .Happy Coding ....
. . .As for the prime number generation, you may want to try Sieve of Eratosthenes... or other Primality tests...

...my favorite, "...most prime numbers from 3 and above obeys ....(6k + 1) or (6k - 1), where k is a natural number"... THat should speed up implementation to some extent....

Thanks for reminding me of Sieve of Eratosthenes. That will work fine for the example test case but not for natural number that is as big as 600851475143, it wont work in java since the largest lenght of array or Set(and Set implementation) you can have is 2^31 (int precision) and Sieve of Eratosthenes reqires boolean flagging over such array.

tohero:
If u mean - will I b using, I wil say php, python nd java. Each at a time

I will b glad to hear ur opinion.

I think you should go with any of C,C++,C#,python or java. Php is a script, I don't think it has such capability of doing trivial algorithmic problem like this. Someone should please correct me if I'm wrong

public class SmallestPositive {

public static void main(String[] arg){
int num, x = 20;
boolean isTrue=false;
for (num = x; !isTrue; num+=x) {
int i=0;
for (i = 2; i <x; i++) {
if(num%i!=0)break;
else if(i==x-1)isTrue=true;
}
if(isTrue)break;
}
System.out.println("smallest positive number that is evenly divisible by all of the numbers from 1 to 20 is "+num);
}

}


smallest positive number that is evenly divisible by all of the numbers from 1 to 20 is 232792560

Yeah, that will make more sense if there is a way to generate palindromes first.
But my implementations generates product n test if it's palindrome from highest to lowest like you said.
Can u post a code or algorithmic representation so I can get you right.

@Kudaisi, you may be bann for 50years for posting this. I guess somebody shouldn't start treaking yet from sokoto, cause moderator still dey sleep now, e never see this post otherwise na 10 minutes to Calaba na en you go see #THREAD CLOSED

[quote author=olyjosh post=33308066]Sorry it not upto a second
Logging Shows its under 20 milli-seconds

tohero:
I can easily formulate the algorithm but writing them with a language is what I am yet to know

Yeah, What language do you use in coding;

lordZOUGA:
How long did this code run?

2 seconds on my 4gb RAM, 1.65GHz processor.
Any better solution without nested loop?

tohero:
Intresting! Unfortunately there are limitations to my contribution

Limitations, How ?

largest palindrome made from the product of two 3-digit numbers.

public class LargestPalindrome2_3 {
static boolean isPalindrome(String s){
return new StringBuffer(s).reverse().toString().equals(s) ;
}

public static void main(String[] args) {
int max = 999;
int min =100;
int prod=0;
boolean isTrue = false;
for (int i = max; i >=min ; i--) {
for (int j = max; j >= min; j--) {
prod=i*j;
if(isPalindrome(String.valueOf(prod))){
isTrue=true;
break;
}
}
if(isTrue)break;
}
System.out.println("largest palindrome made from the product of two 3-digit numbers is:"+prod) ;
}
}



largest palindrome made from the product of two 3-digit numbers is:580085

public class LargestPrimeFactor {
static boolean isPrime(long n){
if(n==1)return false;
if(n==2)return true;
if(n%2==0) return false;
int max = (int)Math.sqrt(n);
for (int i = 2; i <= max; i++) {
if(n%i==0)return false;
}
return true;
}

static Queue<Long> factors(long n){
Queue fac = new ArrayDeque();
//ArrayList<Long> factors = new ArrayList<>();
long h=Math.round(n/2);
for (long i = 2; i <= h; i++) {
if(n%i==0){fac.add(i);}
}
return fac;
}

public static void main(String[] args) {
long lastPrime=0;
long n =600851475143L;
Queue<Long> factors = factors(n);
for (Iterator<Long> iterator = factors.iterator(); iterator.hasNext() {
long next = iterator.next();
if(isPrime(next))lastPrime=next;
}
System.out.println("largest prime factor is: "+lastPrime);
}
}


largest prime factor is: 6857
This solution got me scared cause it runs for about 2minnutes on my 4gb RAM, 1.65GHz processor

Sum of even Fibonacci less than 4000000

public class FibonacciEven {

public static void main(String[] args) {
int num=4000000;
int fib=0;
int t2=2;
int t1=1;
int sum=2;
while (fib<num) {
fib=t1+t2;
t1=t2;t2=fib;
if(fib<num)if(fib%2==0)sum+=fib;
}
System.out.println(sum) ;
}

}


Answer : 4613732

Hi, You can reach me on 08061662025 or olyjoshone@gmail.com so we can talk more on that. I already developed ebook application for client and its hosted on playstore currently

The best thing is to do webservices app. In this approach you will have your mysql online and some script(php, jsp, aps etc) that interact with the database, hence your desktop app call these scripts(via url, HTTP connections). To do this with just java then you will need to do java hosting of your application.

You will need to read on:
deploying javaEE,
deploying RESTFUL Webservice or SOAP

And if you will be using php to querry your database the just try to know how to do RESTFUL webservice in php

Are you getting error? Null pointer I guess. That is beacause variable u is null and you are passing it to float.
What is u in your program

In case you still need this, I come across this website today while I was searching to but I never try it out yet
https://oveem.com.ng/shop/broccoli/

codemarshal08:
+1 for self Training .

+2 for self Training

Guys, una harsh oo. You for ask am say en know whether en no go make hell. That guy woke up in dark and being dried of ideas

What funnies me most is that every other time I use to see thing like, NairaLand need hard working programmer, designer..., then my expectation got rise like coming back another day to say is this for real but seriously every other month I do say is this for real cos I see no changes.

Men, thing are not upto what they were suppose to be here. Like advert is all I see that is new thing here. I expect Nairaland to be one of those thing being discuss probably at silicon valley and most techi blogs/news

Yungwizzzy:
We'll look into it

thanks

That's the only thing I'm tired of about suggestion given to guys at nairaland.
Anyway to my own knowledge, I think Nairaland have very ugly codebase, or because its founded on an open source script that the developers are not flexible with, to change this might be very hard. However, they may need to wait for such implementation.

Please quote me if 'm wrong, and correct me