Problem 38

问题描述：

Take the number 192 and multiply it by each of 1, 2, and 3:

192 * 1 = 192
192 * 2 = 384
192 * 3 = 576

By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)

The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).

What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... , n) where n 1?

解决问题：

public class Problem38 {

	public static boolean IsNumber(int number){
		StringBuffer result = new StringBuffer();
		int mul = 1;
		while(result.length()<9){
			String tmp = number*mul+"";
			result.append(tmp);
			mul ++;
		}
		if(result.length()!=9){
			return false;
		}else{
			boolean[] elements = new boolean[10];
			Arrays.fill(elements, true);
			int r = Integer.parseInt(result.toString());
			System.out.println("Result:"+r);
			while(r!=0){
				int cur = r%10;
				if(cur==0||!elements[cur]){
					return false;
				}
				elements[cur] = false;
				r = r/10;
			}
			return true;
		}
	}
	
	public static void main(String[] args){
		for(int i=9999; i>0; i--){
			if(IsNumber(i)){
				System.out.println(i);
				break;
			}
		}
		System.out.println(IsNumber(9327));
	}
}