Data Structure-Beginning Ability Self-Assessment Questions-4 Have Fun with Numbers (20 points)

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

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

1 #include <stdio.h>
 2  int main()
 3  {
 4      char p[ 21 ];
 5      int a1[ 20 ], a2[ 21 ];     /* Convert the array, a1 is the original, a2 is doubled (considering the carry, more a) */ 
6      int i, j;
 7  
8      scanf( " %s " , p);
 9  
10      /* debug */ 
11  //     printf("%s\n", p); 
12  
13      int n = 0 ;     /* Number of digits */ 
14     for (i = 0 ; i < 20 ; i++ )
 15          if (p[i] != ' \0 ' )
 16          {
 17              a1[i] = p[i] - ' 0 ' ;     /* Character to number */ 
18              a2[i] = a1[i] * 2 ;     /* multiply by 2 */ 
19              n++ ;
 20          }
 21          else 
22          {
 23              break ;
 24          }
 25  
26     /*调试*/
27     //printf("n=%d\n", n);
28     //for (i = 0; i < n; i++)
29     //    printf("%d", a2[i]);
30     //printf("\n");
31 
32     for (j = n - 1; j > 0; j--)    /*进位*/
33         for (i = j; i > 0; i--)
34             if (a2[i] > 9)
35             {
36                 a2[i] -= 10;
37                 a2[i - 1 ] += 1 ;
 38              }
 39  
40      /* Debug */ 
41      // for (i = 0; i < n; i++)
 42      //     printf("%d", a2[i]);
 43      // printf ("\n"); 
44  
45      int t1[ 10 ] = { 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 };     /* Count the number of occurrences */ 
46      int t2[10 ] = { 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 };
 47  
48      /* Debug */ 
49      // for (i = 0; i <10; i++)
 50      //     printf("%d ", t1[i]);
 51      // printf("\n"); 
52  
53      if (a2[ 0 ] > 9 )     /* 1 more digit, No. Same number of digits, continue */ 
54      {
 55          printf( "No\n");
56     }
57     else
58     {
59         for (j = 0; j < 10; j++)
60             for (i = 0; i < n; i++)
61             {
62                 if (a1[i] == j)
63                     t1[j]++;
64                 if (a2[i] == j)
65                     t2[j]++;
66             }
67         for (i = 0; i < 10; i++)     /* compare */ 
68              if (t1[i] == t2[i])
 69                  continue ;
 70              else 
71              {
 72                  printf( " No\n " );
 73                  break ;
 74              }
 75          if (i == 10 )     /* all the same */ 
76              printf( " Yes\n " );
 77      }
 78  
79      /* Debug */ 
80      //for (i = 0; i < 10; i++)
81     //    printf("%d ", t1[i]);
82     //printf("\n");
83     //for (i = 0; i < 10; i++)
84     //    printf("%d ", t2[i]);
85     //printf("\n");
86 
87     for (i = 0; i < n; i++)    /*第二行输出*/
88         printf("%d", a2[i]);
89 
90     return 0;
91 }