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

 1 import java.io.BufferedReader;
 2 import java.io.IOException;
 3 import java.io.InputStreamReader;
 4 
 5 public class number
 6 {
 7     public static void main(String[] args) throws IOException
 8     {
 9         BufferedReader buf=new BufferedReader(new InputStreamReader(System.in));
10         String line=buf.readLine();
11         char[] sou=line.toCharArray();
12         char[] dou;
13         boolean b=false;
14         if(sou[0]-'0'>=5)
15         {
16             b=false;
17             dou=new char[sou.length+1];
18             dou[0]='1';
19             for(int i=sou.length-1,jin=0;i>=0;i--)
20             {
21                 dou[i+1]=(char) (((sou[i]-'0')*2)%10+'0'+jin);
22                 if((sou[i]-'0')*2+jin>=10)
23                     jin=1;
24                 else
25                     jin=0;
26                 
27             }
28         }
29         else
30         {
31             dou=new char[sou.length];
32             for(int i=sou.length-1,jin=0;i>=0;i--)
33             {
34                 
35                 dou[i]=(char) (((sou[i]-'0')*2)%10+'0'+jin);
36                 if((sou[i]-'0')*2+jin>=10)
37                     jin=1;
38                 else
39                     jin=0;
40                 
41             }
42             
43             int[] t1=new int[10];
44             for(int i=0;i<sou.length;i++)
45             {
46                 t1[sou[i]-'0']++;
47             }
48             int[] t2=new int[10];
49             for(int i=0;i<dou.length;i++)
50             {
51                 t2[dou[i]-'0']++;
52             }
53             for(int i=0;i<t1.length;i++)
54             {
55                 if(t1[i]!=t2[i])
56                 {
57                     b=false;
58                     break;
59                 }
60                 b=true;
61             }
62         }
63 
64         if(b==true)
65         {
66             System.out.println("Yes");
67             System.out.println(dou);
68         }
69         else
70         {
71             System.out.println("No");
72             System.out.println(dou);
73         }
74             
75         
76     }
77 }