[Digital DP] HDU 3967 Zero's numberd

For two numbers I and K, divide them into two parts: Number, N and M. If (n + M) % K = 0, this is a method.

For example, 333 can be divided into 3, 33, 33, and 3. For (333,3), it is equal to 2.

A, B, and K are given, which is (~ B, k) How many methods are there?

Idea: For a number, pay attention to the leading zero and enumerate the points.

DP [22] [22] [22] [22] [2], representing the I-bit, the Shard is fd, and the remainder mod. For K remainder, whether there is a leading zero

Code:

# Include "cstdlib" # include "cstdio" # include "cstring" # include "cmath" # include "stack" # include "algorithm" # include "iostream" using namespace STD; __int64 DP [22] [22] [22] [22] [2], ten [22]; // I bit, the Shard is fd, the remainder mod, for K remainder, whether there is a leading zero int num [22], CNT ;__ int64 DFS (INT site, int FD, int mod, int K, int ff, int F) {If (Site = 0) return MoD? 0: 1; if (! F & DP [site] [FD] [mod] [k] [ff]! =-1) return DP [site] [FD] [mod] [k] [ff]; int Len = f? Num [site]: 9; _ int64 ans = 0; For (INT I = 0; I <= Len; I ++) {If (site> FD) {If (FF = 0 & Site = FD + 1 & I = 0) continue; // The leading 0 cannot be before the shard, that is, the first half cannot be 0 ans + = DFS (site-1, FD, (I * Ten [site-FD] + mod) % K, K, FF | I! = 0, F & I = Len);} else ans + = DFS (site-1, FD, (I * Ten [site] + mod) % K, K, FF | I! = 0, F & I = Len);} If (! F) DP [site] [FD] [mod] [k] [ff] = ans; return ans; }__ int64 solve (_ int64 X, int K) {CNT = 0; If (x <10) return 0; _ int64 ans = 0; while (x) {num [++ CNT] = x % 10; x/= 10;} For (INT I = 1; I <CNT; I ++) // after the division of enumeration points is ans + = DFS (CNT, I, 0, k, 0, 1); Return ans;} int main () {_ int64 a, B; int K; memset (DP,-1, sizeof (DP )); ten [1] = 1; for (INT I = 2; I <= 18; I ++) Ten [I] = ten [I-1] * 10; while (scanf ("% i64d % i64d % d", & A, & B, & K )! =-1) {printf ("% i64d \ n", solve (B, k)-solve (A-1, k);} return 0 ;}

[Digital DP] HDU 3967 Zero's numberd