Java on Rohit Jha

0x000c - Project Euler 010

Wed, 02 Nov 2011 12:08:12 +0000

Problem:

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

Find the sum of all the primes below two million.

Solution:

1. Brute-force approach:


class PE010_brute
{
 static boolean isPrime(long n)
 {
 if(n == 2 || n == 3) return true;
	if(n < 2 || (n % 2 == 0)) return false;
 if(n < 9) return true;
 if(n % 3 == 0) return false;

 int r = (int)Math.sqrt(n), f = 5;

 while(f <= r) {
 if ((n % f == 0) || (n % (f + 2) == 0))
 return false;
 f += 6;
 }
 return true;
 }
}

2. Efficient approach (using the Sieve of Eratosthenes) :

0x0002 - Project Euler 002 – More efficient

Sun, 25 Sep 2011 18:42:33 +0000

Problem:

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

Solutions:

1. Python Solution:


limit = 4000000
x, y, sum = 3, 5, 2

while sum < limit:
 sum += (x + y)
 x, y = x + 2*y, 2*x + 3*y

print(sum)

2. Java Solution:

0x0001 - Project Euler 002

Wed, 07 Sep 2011 19:18:01 +0000

Problem:

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

Solutions:

1. Python solution:


from math import sqrt

limit, current, sum = 4000000, 2, 0
phi3 = ((1+sqrt(5))/2)**3

while sum < limit:
 sum += current
 current = round(phi3*current)

print (sum)

2. Java solution:

0x0000 - Project Euler 001

Sat, 03 Sep 2011 17:14:31 +0000

Problem:

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

Solutions: 1. Python solution:


def sum_n(n, d) :
 n = int(n/d)
 return int(d * n * (n + 1) / 2)

print (sum_n(999, 3) + sum_n(999, 5) - sum_n(999, 15))

2. Java solution:


class PE001
{
	public static int sum_n(int n,int d)
	{
		n = (int)(n/d);
		return (int)(d * n * (n + 1) / 2);
	}

	public static void main(String args[])
	{
		System.out.println(sum_n(999, 3) + sum_n(999, 5) - sum_n(999, 15));
	}
}

3. Perl solution: