Problem 009 : Special Pythagorean Triplet

Special Pythagorean triplet

Problem 9

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

a² + b² = c²

For example, 3² + 4² = 9 + 16 = 25 = 5².
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.

                                                                                                           https://projecteuler.net/problem=9

Problem 9 is about Pythagorean triplet.

The basic formula is :

A² + B² = C²

Using this formula will easily solve the problem.

One quick trick is to set the limit to int A and B so that we can reduce the time.

Assuming that B is always greater than or equal to A, B will always start with A.

Also, A will be less than one third of the perimeter since A is the smallest side of a triangle.

B will be less than one half of the perimeter since B should be smaller than the hypotenuse, C.

for (int a=1, end = 1000/3; a < end; a++)
{
 for (int b=a, end2 = 1000/2; b < end2; b++)
 {
  int c = 1000-a-b;
  if (c>0 && a*a + b*b == c*c)
  {
   System.out.println(a*b*c);
   return;
  }
 }
}

Answer is 31875000

Execution time is 1.7 ms, or 0.0017 seconds

Source code: https://github.com/Ainodyne/Project-Euler/blob/master/Problem009.java