public static String solution(int n){

StringBuffer sb = new StringBuffer();

int wei = (int)(Math.log(n+1)/Math.log(2)); // 求出第n个幸运数的位长度

int num = n - ((int)Math.pow(2,wei)-1); // 求出第n个幸运数在该层的位置值

String binNum = Integer.toBinaryString(num);

for(int i = 0 ; i < wei-binNum.length() ; i++)

{

sb.append("0");

}

sb.append(binNum);

String luckyNum = sb.toString();

luckyNum = luckyNum.replace("0", "4");

luckyNum = luckyNum.replace("1", "7");

return luckyNum;

	}