1023 Have Fun with Numbers (20 分)

1023 Have Fun with Numbers (20 分)

 

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 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<vector>
 #include<queue>
 #include<stack>
 #include<algorithm>
 #include<string>
 using namespace std;
 int Array[];
 int main()
 {
     string num;
     cin >> num;
     int len = num.size();
     int flag = ;
     int temp = ;
     for (int i = len - ; i >= ; i--)
     {
         Array[num[i] - '']++;
         temp = (num[i] - '') *  + flag;
         flag = ;
         if (temp> )
         {
             temp %= ;
             flag = ;
         }
         Array[temp]--;
         num[i] = '' + temp;
     }
     int flag1 = ;
     for(int i=;i<;i++)
         if (Array[i]!=)
         {
             flag1= ;
             break;
         }
     if (flag1)
     {
         cout << "No" << endl;
         if (flag)
             cout << "" << num;
         else
             cout << num;
     }
     else
     {
         cout << "Yes" << endl;
         if (flag)
             cout << "" << num;
         else
             cout << num;
     }
 
 }