C-search for the number of background qualities

I saw this problem when I checked my interview experience,

First, Glossary:

The number of input records is the same as that of the last query, for example, 121,12321.

Prime Number: a product can only be divided by 1 and itself.

The number of input objects: that is, the number of input objects and the number of input objects.

Okay. Now that the concept is fully explained, let's write it.ProgramNow.

 1 # Include <stdio. h> 2 # Include <stdlib. h> 3 # Include <math. h> 4   /*  *  5   Determines whether M is a prime number.  6   @ Return 1: prime number, 0: not a prime number  7  *  */  8   Int SS ( Int  M ){  9       For ( Int I = 2 ; I <= SQRT (m); I ++ )  10           If (! (M % I ))  11               Return  0  ;  12       Return   1  ;  13   }  14   15   /*  *  16   Determine whether M is the number of replies  17   @ Return 1: return, 0: not return  18   * */  19   Int Huiwen ( Int  M ){  20       Int X = 0  , Y;  21 Y = M;  22       While  (Y ){  23 X = x * 10 + Y %10  ;  24 Y/= 10  ;  25   }  26       If (M = X)  27           Return   1  ;  28       Else  29          Return   0  ;  30   }  31   32   Int  Main ()  33   {  34       /*  *  35   The requirements for the number of whitelists:  36   1-digit must be odd 37   2. Number of replies + Prime Number  38   3. Only Exception: 11  39   *  */  40       //  Print 11 Directly  41 Printf ( "  The number of whitelists: % d \ r \ n  " , 11  );  42      Int J = 1 ; //  Base  43       Int K = 3 ; //  Boundary  44       Int  L;  45       While (J < K ){  46 L = POW (100  , J );  47           For ( Int I = L + 1 ; I < 10 * L; I ++, I ++ )  48               If (Huiwen (I )&& SS (I ))  49 Printf ( "  The number of whitelists: % d \ r \ n  " , I );  50 J ++ ;  51   }  52 }

 

Why do we need odd numbers of replies? Because the number of even bits is destined to be divisible by 11, it is definitely not a prime number.

So here we only need to consider 101-999,10001-99999, and so on odd digits, so we can reduce a lot of loops.