Acm hdu 1282 echo number conjecture (simple question)

Search Result Conjecture

Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 1524 accepted submission (s): 945

Problem description is a positive integer. If the read from left to right (called positive order number) is the same as the read from right to left (called reverse order number), such a number is called the return number. Take a positive integer. If it is not a return number, add the number to its reverse order number. If it is not a return number, repeat the above steps until the number is obtained. For example, 68 is changed to 154 (68 + 86), then to 605 (154 + 451), and finally to 1111 (605 + 506), while 1111 is the number of replies. Therefore, a mathematician proposed a conjecture: No matter what positive integer is at the beginning, after a finite ordinal number and a descending order number are added, a return number is obtained. So far, I do not know whether this conjecture is true or false. Compile Program Verify.

Each input row has a positive integer.
Note: ensure that the intermediate result of the input data is less than 2 ^ 31.

Output corresponds to each input. Two rows are output, one is the number of transformations, and the other is the transformation process.

Sample input27228
37649

Sample output3
27228 ---> 109500 -----> 115401 ---> 219912
2
37649 ---> 132322 ---> 355553

Authorsmallbeer (CRF)

Source hangdian ACM training team training competition (VII)

Recommendlcy

 # Include  <  Stdio. h  >  
  Int  Change (  Int  N)
{
  Int A [  20  ];
  Int  K  =  0  ;
  While  (N  ! =  0  )
{
K  ++  ;
A [k]  =  N  % 10  ;
N  /=  10  ;
}
  Int  Res  =  0  ;
  For  (  Int  I  =  1  ; I  <= K; I  ++  )
{
Res  * =  10  ;
Res  + =  A [I];
}
  Return  Res;
}
  Int  Main ()
{
  Int  N, I, CNT;
  Int  Result [ 100  ];
  While  (Scanf (  "  % D  "  ,  &  N)  ! =  EOF)
{
CNT  =  0  ;
Result [  0  ] =  N;
  While  (N  ! =  Change (n ))
{
N  =  N  +  Change (N );
Result [  ++  CNT]  =  N;
}
Printf (  "  % D \ n "  , CNT );
  For  (I  =  0  ; I  <  CNT; I  ++  )
Printf (  "  % D --->  "  , Result [I]);
Printf (  "  % D \ n "  , Result [CNT]);
}
  Return     0  ;
}