[Codeforces 258b & 259 D] little elephant and elections digital DP + DFS

The numbers 4 and 7 in the numbers selected by the Party specified by the queue (different) are more than the sum of the numbers owned by the other six parties. StrictMultiple, number of inquiry solutions. For example, when M = 7, the remaining random options will have at least one 4 or 7, which is in conflict with the question. Therefore, when the number of solutions is 0 m = 8, 71 2 3 5 6 8 is a legal solution
Idea: because the number 4 and 7 selected by the elephant's party contains at most the same number as the number of digits m, the number of 4 and 7 contained in the elephant's party is enumerated, use DFS directly for the remaining 6 Parties (using the multiplication principle ). The digital DP can be used to calculate 1 ~ M contains the number of K lucky numbers. Here, we use a memory-based search method for ease of thinking. For details, see the code and comments.
Code:

# Include <cstdio> # include <cstring> # include <cmath> # include <map> # include <algorithm> # include <cstdlib> # define sizes 40000 # define mod 1000000007 llusing namespace STD; long long DP [20] [20]; long res [20]; int bit [20], Bn; long m, n, Pn, ans; void deep (INT now, int _ max, int party, long cur) // now indicates the number of lucky numbers, _ max limit, and Party indicates the serial number, cur is the solution count {If (now> _ max) return; // If (Party = 7) {If (now <_ max) {ans + = C Ur; // Add ans % = MOD;} return ;}for (INT I = 0; I <Bn; I ++) {If (RES [I]) // is it feasible? {res [I] --; deep (now + I, _ max, party + 1, cur * (RES [I] + 1) % mod ); res [I] ++ ;}}long long DFS (long POs, long target, long limit) // POS indicates the number of digits processed, target requires a few lucky numbers. Limit indicates whether there is a limit {long sum = 0; If (Pos = 0) return target = 0; // If (Limit = 0) & (DP [POS] [target]! =-1) return DP [POS] [target]; // long tail = limit? Bit [POS]: 9; // determine the upper limit of enumeration for (INT I = 0; I <= tail; I ++) sum + = DFS (pos-1, target-(I = 4 | I = 7), (Limit = 1) & (bit [POS] = I )); if (Limit = 0) DP [POS] [target] = sum; return sum;} void CAL (long x) {ans = Bn = 0; long long y = x; while (y) {++ Bn; bit [bn] = Y % 10; y/= 10;} For (INT I = 0; I <= Bn; I ++) RES [I] = DFS (bn, I, 1); // calculate the number of optional res in the memory search [0]-= 1; // remove the for (INT I = 1; I <= Bn; I ++) deep (0, I, 1, Res [I]) where a single number is 0; // enumerate the number of lucky numbers} int main () {long m; memset (DP,-1, sizeof DP); scanf ("% i64d", & M ); cal (m); printf ("% i64d \ n", ANS); Return 0 ;}