Problem:
Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
Triangle Tn=n(n+1)/2 1, 3, 6, 10, 15, ...
Pentagonal Pn=n(3n[−]1)/2 1, 5, 12, 22, 35, ...
Hexagonal Hn=n(2n[−]1) 1, 6, 15, 28, 45, ...
It can be verified that T285 = P165 = H143 = 40755.
Find the next triangle number that is also pentagonal and hexagonal.
Solution:
1533776805
Code:
The solution may include methods that will be found here: Library.java .
public interface EulerSolution{
public String run();
}
/*
* Solution to Project Euler problem 45
* By Nayuki Minase
*
* http://nayuki.eigenstate.org/page/project-euler-solutions
* https://github.com/nayuki/Project-Euler-solutions
*/
public final class p045 implements EulerSolution {
public static void main(String[] args) {
System.out.println(new p045().run());
}
public String run() {
int i = 286;
int j = 166;
int k = 144;
while (true) {
long triangle = (long)i * (i + 1) / 2;
long pentagon = (long)j * (j * 3 - 1) / 2;
long hexagon = (long)k * (k * 2 - 1);
long min = Math.min(Math.min(triangle, pentagon), hexagon);
if (min == triangle && min == pentagon && min == hexagon)
return Long.toString(min);
if (min == triangle) i++;
if (min == pentagon) j++;
if (min == hexagon ) k++;
}
}
}
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