project euler 119

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本文探讨了一种特殊的数列，其中每个数等于其各位数字的幂次和。通过编程方式寻找该数列的第30项。利用Java实现算法，通过遍历可能的基数与指数组合来找到符合条件的数字。

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Digit power sum

The number 512 is interesting because it is equal to the sum of its digits raised to some power: 5 + 1 + 2 = 8, and 83 = 512. Another example of a number with this property is 614656 = 284.

We shall define an to be the nth term of this sequence and insist that a number must contain at least two digits to have a sum.

You are given that a2 = 512 and a10 = 614656.

Find a30.

package projecteuler;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;

import junit.framework.TestCase;

public class Prj119 extends TestCase {

	public void testDigitPowerSum() {
		
		calAofPowerB();
	}

	long sumDigit(long val) {

		long v = 0;
		String str = Long.toString(val);
		for (int i = 0; i < str.length(); i++) {
			v += Long.parseLong(String.valueOf(str.charAt(i)));
		}
		return v;
	}
	
	void calAofPowerB(){
		List<Long> dataList = new ArrayList<Long>();
		for( int a = 2 ; a <= 100; a ++){
			for( int b = 2; b <= 64; b ++){
				Long v = (long) Math.pow(a, b);
				if( sumDigit(v) == a){
					dataList.add(v);
				}
			}
		}
		
		Collections.sort(dataList);
		System.out.println(Arrays.toString(dataList.toArray()));
		System.out.println(dataList.size());
		System.out.println(dataList.get(29));
	}

}