Java的大数计算BigNumber

Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. Check to see the result if we double it again!

Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.

Input Specification:

Each input file contains one test case. Each case contains one positive integer with no more than 20 digits.

Output Specification:

For each test case, first print in a line "Yes" if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or "No" if not. Then in the next line, print the doubled number.

Sample Input:

1234567899

Sample Output:

Yes
2469135798

不知道为什么有一个样例没过，觉得写得没毛病啊

 import java.math.BigInteger;
 import java.util.Scanner;

 public class A {

     public static void main(String[] args) {
         Scanner cin = new Scanner(System.in);
         BigInteger a = cin.nextBigInteger();
         BigInteger bb = a;
         BigInteger c = BigInteger.valueOf(10);
         BigInteger d;
         BigInteger ans = a.add(bb);
         BigInteger b = ans;
         boolean book = true;
         int[] book0 = new int[10];
         while(!a.equals(BigInteger.valueOf(0))) {
 //            System.out.println(a.equals(0));
             d = a.mod(c);
             a = a.divide(c);
             book0[d.intValue()] = 1;
         }
         while(!b.equals(BigInteger.ZERO)) {
             d = b.mod(c);
             b = b.divide(c);
             if(book0[d.intValue()] != 1) {
                 book = false;
                 break;
             }
         }
         if(book == false) {
             System.out.println("No");
         }
         else {
             System.out.println("Yes");
         }
         System.out.println(ans);

     }

 }