hdu3925 A至少加上多少包含B（java）

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从a的最低位开始枚举和B比较。
import java.math.BigInteger;import java.util.Scanner;public class Main{    public static void main(String[] args)    {        Scanner cin = new Scanner(System.in);        BigInteger zero = BigInteger.valueOf(0);        BigInteger ten = BigInteger.valueOf(10);        int i, j, ncase;        BigInteger a, b, min;        ncase = cin.nextInt();        for(j=1;j<=ncase;j++)        {            min = new BigInteger("99999999999999999999999999");            a = cin.nextBigInteger();            b = cin.nextBigInteger();            if (a.toString().indexOf(b.toString()) != -1)//a包含b            {                min = zero;            }            else            {                BigInteger tmp1, tmp2;                for (i = b.toString().length(); i <= 105; i++)                {                    tmp1 = a.mod(ten.pow(i));//a的后几位                    if(tmp1.compareTo(b)==0) 后几位和b相等                    {                        min=zero;                        break;                    }                    else if(tmp1.compareTo(b)==-1)//小于b                    {                        tmp2 = b.subtract(tmp1);                        if(tmp2.compareTo(min)==-1)                        {                            min=tmp2;                        }                    }                    else //大于b，在b的前面加1                    {                        tmp2 = b.add(ten.pow(b.toString().length()));                        tmp2 = tmp2.subtract(tmp1);                        if (tmp2.compareTo(min) == -1&& tmp2.compareTo(zero) != -1)                        {                            min = tmp2;                        }                    }                    b = b.multiply(ten);//b乘以10                }                            }            System.out.println("Case #" + j +": " + min);        }    }}