HDU 3709 balanced number zoj 3416 balanced number (Digital DP)

Balanced number
Time limit:5 seconds Memory limit:65536 KB

A balanced number is a non-negative integer that can be balanced if a packet is placed at some digit. more specifically, imagine each digit as a box with weight indicated by the digit. when a distance is placed at some digit of the number, the distance from a digit to the distance is the offset between it and the distance. then the torques of left part and right part can be calculated. it is balanced if they are the same. A balanced number must be balanced with the specified at some of its digits. for example, 4139 is a balanced number with fixed at 3. the torqueses are 4*2 + 1*1 = 9 and 9*1 = 9, for left part and right part, respectively. it's your job to calculate the number of balanced numbers in a given range [X,Y].

Input

The input contains multiple test cases. The first line is the total number of casesT(0 <T≤ 30). For each case, there are two integers separated by a space in a line,XAndY. (0 ≤X≤Y≤ 1018 ).

Output

For each case, print the number of balanced numbers in the range [X,Y] In a line.

Sample Input

 
20 97604 24324
 

Sample output

 
10897
 

Author:Gao, Yuan
Source:The 2010 ACM-ICPC Asia Chengdu Regional Contest

 

 

 

A good number of DP questions.

The value must be the number of balances. One Dimension is added to indicate the torque.

/*  * HDU 3709 * balanced number, enumeration pivot * DP [I] [J] [k] I indicates the number of processed digits, J indicates the pivot, and K indicates the torque and  */  # Include <Iostream> # Include <Stdio. h> # Include <Algorithm> # Include < String . H> Using   Namespace  STD;  Long   Long DP [ 20 ] [ 20 ] [ 2000  ];  Int Bit [ 20  ];  Long   Long DFS ( Int POs, Int Center, Int PRE, Bool  Flag ){  If (Pos =- 1 )Return Pre = 0  ;  If (Pre < 0 ) Return   0 ; //  Current Torque is negative, pruning      If (! Flag & DP [POS] [center] [pre]! =- 1  )  Return  DP [POS] [center] [pre];  Int End = flag? Bit [POS]: 9  ;  Long   Long Ans = 0  ;  For ( Int I = 0 ; I <= end; I ++ ) Ans + = DFS (POS- 1 , Center, pre + I * (POS-center), flag & I = End );  If (! Flag) DP [POS] [center] [pre] =Ans;  Return  Ans ;}  Long   Long Calc ( Long   Long  N ){  Int Len = 0  ;  While  (N) {bit [Len ++] = N % 10  ; N /= 10 ;}  Long   Long Ans = 0  ;  For ( Int I = 0 ; I <Len; I ++ ) Ans + = DFS (LEN- 1 , I, 0 , 1  );  Return Ans-(LEN- 1 ); //  Remove all zeros  }  Int  Main (){  //  Freopen ("in.txt", "r", stdin );  //  Freopen ("out.txt", "W", stdout );      Int  T;  Long   Long  X, Y; memset (DP, - 1 ,Sizeof (DP )); //  Do not forget this initialization. Scanf ( "  % D  " ,& T );  While (T -- ) {Scanf (  "  % I64d % i64d  " , & X ,& Y); printf (  "  % I64d \ n " , Calc (y)-Calc (X- 1  ));}  Return   0  ;}