UESTC 1307 windy (Digital DP)

Question link: http://acm.uestc.edu.cn/problem.php? PID = 1, 1307

 

Windy number time limit: 1000 MS Memory limit: 65536 KB Solved: 104 Tried: 720 Submitstatusbest solutionbackdescription

Windy defines the number of windy.
A positive integer that does not contain leading zero and the difference between two adjacent numbers is at least 2 is called the windy number.
Windy: How many windy numbers are there between A and B, including a and B?

Input

Contains two integers, a B.
1 <= A <= B <= 2000000000.

Output

Contains an integer: the number of windy numbers in the closed interval [a, B.

Sample Input

1 10

Sample output

9

Source

Windy

For a simple digital DP, the difference between two adjacent numbers must be greater than or equal to 2.

 /*  * UESTC 1307 * Number of windy: Excluding leading 0, the difference between two adjacent numbers is at least a positive integer of 2 * calculate the number of windy between [a, B, 1 <= A <= B <= 2000000000  */  # Include <Iostream> # Include < String . H> # Include <Stdio. h> # Include <Algorithm> Using   Namespace  STD;  Int DP [15 ] [ 10 ]; //  DP [I] [J] indicates the number of windy numbers whose length is I and whose highest bit is J.  Void  Init () {memset (DP,  0 , Sizeof  (DP ));  For ( Int I = 0 ; I < 10 ; I ++ ) DP [  1 ] [I] = 1  ;  For ( Int I = 2 ; I <= 10 ; I ++ )  For ( Int J = 0 ; J < 10 ; J ++ ){  For ( Int K =0 ; K <= J- 2 ; K ++ ) DP [I] [J] + = DP [I- 1  ] [K];  For ( Int K = J + 2 ; K < 10 ; K ++ ) DP [I] [J] + = DP [I- 1  ] [K] ;}}  Int Bit [ 20 ];  Int Calc ( Int  N ){  If (N = 0 ) Return   0  ;  Int Len = 0  ;  While  (N) {bit [ ++ Len] = n % 10  ; N /= 10  ;} Bit [Len + 1 ] =- 10  ;  Int Ans = 0  ;  Bool Flag = True  ;  For ( Int I = 1 ; I <Len; I ++) // Calculate the number of characters whose length is less than Len          For ( Int J = 1 ; J <= 9 ; J ++ ) Ans + = DP [I] [J];  For ( Int J = 1 ; J <bit [Len]; j ++) //  Highest bit Ans + = DP [Len] [J];  For (Int I = len- 1 ; I> = 1 ; I -- ){  For ( Int J = 0 ; J <bit [I]; j ++ )  If (ABS (bit [I + 1 ]-J)> = 2  ) Ans + = DP [I] [J];  If (ABS (bit [I + 1 ]-Bit [I]) < 2  ) {Flag = False  ;  Break  ;}}  If (FLAG) ans ++ ;  Return  Ans ;}  Int  Main (){  //  Freopen ("in.txt", "r", stdin ); //  Freopen ("out.txt", "W", stdout );      Int  A, B; Init ();  While (Scanf ( "  % D  " , & A, & B) = 2  ) {Printf (  "  % D \ n  " , Calc (B)-Calc (- 1  ));} Return   0  ;}