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:

Yes2469135798

#include
      
       #include
       
        #include 
        
         using namespace std;int sort(int a[],int n){    int temp;    for(int i=0;i
         
          a[j]){                temp=a[i];                a[i]=a[j];                a[j]=temp;            }        }            }    return 0;}int main(){    string num;    stringstream ss;    int size,j=0;    cin>>num;    size=num.size();    int *a=new int[size];    int *b=new int[size+1];    int *doubleNum=new int[size+1];    for(int i=0;i
          
           0;i--){        if(a[i-1]+a[i-1]>=10){            doubleNum[i]+=(a[i-1]+a[i-1])%10;            doubleNum[i-1]+=1;        }else{            doubleNum[i]+=a[i-1]+a[i-1];            }                }    if(doubleNum[0]==0){        for(int i=0;i
          
         
        
       
      

测试

结果