00-自测4. Have Fun with Numbers (20)

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

---

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class number
{
    public static void main(String[] args) throws IOException
    {
        BufferedReader buf=new BufferedReader(new InputStreamReader(System.in));
        String line=buf.readLine();
        char[] sou=line.toCharArray();
        char[] dou;
        boolean b=false;
        if(sou[0]-'0'>=5)
        {
            b=false;
            dou=new char[sou.length+1];
            dou[0]='1';
            for(int i=sou.length-1,jin=0;i>=0;i--)
            {
                dou[i+1]=(char) (((sou[i]-'0')*2)%10+'0'+jin);
                if((sou[i]-'0')*2+jin>=10)
                    jin=1;
                else
                    jin=0;
                
            }
        }
        else
        {
            dou=new char[sou.length];
            for(int i=sou.length-1,jin=0;i>=0;i--)
            {
                
                dou[i]=(char) (((sou[i]-'0')*2)%10+'0'+jin);
                if((sou[i]-'0')*2+jin>=10)
                    jin=1;
                else
                    jin=0;
                
            }
            
            int[] t1=new int[10];
            for(int i=0;i<sou.length;i++)
            {
                t1[sou[i]-'0']++;
            }
            int[] t2=new int[10];
            for(int i=0;i<dou.length;i++)
            {
                t2[dou[i]-'0']++;
            }
            for(int i=0;i<t1.length;i++)
            {
                if(t1[i]!=t2[i])
                {
                    b=false;
                    break;
                }
                b=true;
            }
        }

        if(b==true)
        {
            System.out.println("Yes");
            System.out.println(dou);
        }
        else
        {
            System.out.println("No");
            System.out.println(dou);
        }
            
        
    }
}