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"); } }