Have Fun with Numbers

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<stdio.h>
 #include<string.h>
 int hash1[]={};
 int hash2[]={};

 int jud(char a[],int len)
 {
     if(a[]>=''&&a[]<='')
     {
         return ; //false代表不合格的进位大整数；
     }
     if(a[]==''&&len!=)
     {
         return ;
     }
         if(a[]==''&&len==)
     {
         return ;
     }
     if(a[]>=''&&a[]<'')
     {
         return ;
     }
 }
 int main()
 {
     char a[];
     int inta[],i,len,temp;
     gets(a);
     len=strlen(a);
     /*将字符串数组转化为对应整数，存入变换的整数数组中*/
     for(i=;i<len;i++)
     {
         inta[i]=a[i]-'';
         hash1[inta[i]]++;
     }
     /*/检测输出
         for(i=0;i<len;i++)
     {
         printf("%d",inta[i]);
     }
     */
     //检测变化数组内容的数字出现次数；
     /**/
 //        printf("\n");
 //        for(i=0;i<10;i++)
 //    {
 //        printf("%d ",hash1[i]);
 //    }
     /**/
 //    printf("\n");
      int count=;
     for(i=len-;i>;i--)
     {
         temp=inta[i]*+count;
         if(temp>=)
         {
             count=;
         }
         else{
             count=;
         }
         inta[i]=temp%;
     }
     inta[]=inta[]*+count;
     //将变换数组*2

     //判断数组
 if(    jud(a,len)==)
 {

 //对*2后的数组中数字出现的次数进行统计；
         for(i=;i<len;i++)
     {
         hash2[inta[i]]++;
     }
 //        for(i=0;i<10;i++)
 //    {
 //        printf("%d ",hash2[i]);
 //    }

     //比较两hash表是否完全相等
     for(i=;i<;i++)
     {
         if(hash1[i]!=hash2[i])
         {
             printf("No\n");
                 for(i=;i<len;i++)
             {
                 printf("%d",inta[i]);
              }
             return ;
         }
     }
         printf("Yes\n");
         for(i=;i<len;i++)
     {
         printf("%d",inta[i]);
      }
 }

 else if(jud(a,len)==)
 {
         printf("No\n");
         for(i=;i<len;i++)
         {
             printf("%d",inta[i]);
         }
 }
 else
 {
     printf("No\n");
             for(i=;i<len;i++)
         {
             printf("%d",inta[i]);
         }
 }

     return ;
  }