Have Fun with Numbers (大数)

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

 #include <iostream>

 #include <string>

 #include <algorithm>

 using namespace std;

 int aa[];

 int v1[];

 int v2[];

 int main()

 {

       string  ss;

     while(cin>>ss)

       {

           int i;

             for(i=;i<;i++)

             {

                 aa[i]=;

             }

             for(i=;i<;i++)

             {

               v1[i]=;

               v1[i]=;

             }

             int count=;

             for(i=ss.length()-;i>=;i--)

             {

                aa[count++]=ss[i]-'';

                v1[ss[i]-'']=;

             }

             for(i=;i<count;i++)

             {

                 aa[i]=aa[i]*;

             }

           int tem,len;

         for(i=;i<count;i++)

             {

                 if(aa[i]>)

                   {

                      tem=aa[i]/;

                      aa[i+]=aa[i+]+tem;

                      aa[i]=aa[i]%; 

                   }

             }

             if(aa[count]==) len=count;

             else len=count+;

         reverse(aa,aa+len);

         for(i=;i<len;i++)

         {

            v2[aa[i]]=;

         }

       bool ifis=true;

         for(i=;i<;i++)

         {

            if(v1[i]!=v2[i]) ifis=false;

         }

         if(ifis) cout<<"Yes"<<endl;

         else  cout<<"No"<<endl;

         for(i=;i<len;i++)

         {

            cout<<aa[i];

         }

         cout<<endl;

       }

       return ;

 }