Problem44

原题：

Pentagonal numbers are generated by the formula, Pn=n(3n-1)/2.
 The first ten pentagonal numbers are:

1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...

It can be seen that P4 + P7 = 22 + 70 = 92 = P8.
However, their difference, 70-22 = 48, is not pentagonal.

Find the pair of pentagonal numbers, Pj and Pk,
for which their sum and difference is pentagonal and D = |Pk-Pj| is minimised;
what is the value of D?

 

代码如下：

public class Problem44 {//存储pentagonal数的集合。static Set<Integer> pentagonalList = new TreeSet<Integer>();static int getIsPentagonal(int n){return n*(3*n-1)/2;}//该数和集合中的数进行运算，看是否满足条件。//当前的number为pentagonal数，pentagonalList集合中的数位pentagonal数，//只要保证number-i与i的绝对值之差为pentagonal数，即为所求static boolean getTheResult(int number){for(int i:pentagonalList){if(pentagonalList.contains(Integer.valueOf(number-i))&&i!=number-i&&pentagonalList.contains(Math.abs(number-2*i))){System.out.println("this number1 :"+i);System.out.println("this number2 :"+(number-i));System.out.println("| number2-number1 |:"+(Math.abs(number-2*i)));return true;}}return false;}public static void main(String[] args) {long beginTime = System.currentTimeMillis();int n = 1;while(true){int currentNumber = getIsPentagonal(n);//System.out.println("Number:"+currentNumber);pentagonalList.add(currentNumber);n++;if(getTheResult(currentNumber)){System.out.println("| number2+number1 |:"+currentNumber);break;}}long endTime = System.currentTimeMillis();System.out.println(endTime-beginTime+"ms");}}