Problem 45

問題描述：

 

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(3n1)/2		1, 5, 12, 22, 35, ...
Hexagonal		Hn=n(2n1)		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.

 

 

解決問題：

 

 

package projecteuler;

import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;

public class Problem45 {

	public static void main(String[] args){

		Map<Long, Long> number = new HashMap<Long, Long>();
		
		Long Triangle ;
		Long Pentagonal ;
		Long Hexagonal ;
		Long T = 286L;
		boolean find = false;
		do{
			Triangle = (T*(T+1))/2L;
			Pentagonal = (T*(3*T-1))/2L;
			Hexagonal = T*(2*T-1L);
			
			number.put(Hexagonal, 2L);
			if(number.containsKey(Triangle)){
				Long value = number.get(Triangle);
				if(value==1){
					find = true;
					System.out.println("Triangle:"+Triangle);
				}else{
					number.remove(Triangle);
					number.put(Triangle, 1L);
				}
			}
			if(number.containsKey(Pentagonal)){
				Long value = number.get(Pentagonal);
				if(value==1){
					find = true;
					System.out.println("Pentagonal:"+Pentagonal);
				}else{
					number.remove(Pentagonal);
					number.put(Pentagonal, 1L);
				}
			}
			System.out.println("T:"+T);
			T++;
		}while(!find);
		long r = T*(T+1)/2L;
		System.out.print(r);
	}
}