CC favnum (AC automation + digital DP)

Reprinted please indicate the source, thank youHttp://blog.csdn.net/ACM_cxlove? Viewmode = Contents
By --- cxlove

The question that Xiaodao gave, codechef for the first time, looks very high.

Some pattern strings are provided. The specified pattern string contains at least one pattern string.

The K specified string in the question interval.

Add the pattern string to the automatic machine, and then obtain the number of specified strings in the interval.

Just answer the second question.

In fact, it's still very watery...

Prepare to follow the suffix of Xiaodao to learn the automatic machine.

# Include <iostream> # include <cstdio> # include <map> # include <cstring> # include <cmath> # include <vector> # include <algorithm> # include <set> # include <string> # include <queue> # define INF 100000005 # define M 20005 # define n 2000 # define maxn 300005 # define EPS 1e-10 # define zero () FABS (a) <EPS # define min (A, B) (a) <(B )? (A) :( B) # define max (A, B) (a)> (B )? (A) :( B) # define Pb (a) push_back (a) // # define MP (a, B) make_pair (a, B) # define MEM (, b) memset (a, B, sizeof ()) # define ll long # define mod 1000000007 # define lson step <1 # define rson step <1 | 1 # define sqr (A) (a) * ()) # define key_value ch [CH [root] [1] [0] // # pragma comment (linker, "/Stack: 1024000000,1024000000") using namespace STD; struct trie {trie * Next [10]; trie * fail; int isword, kind;}; trie * que [m], s [m]; Int idx; trie * newnode () {trie * TMP = & S [idx]; MEM (TMP-> next, null); TMP-> isword = 0; TMP-> fail = NULL; TMP-> kind = idx ++; return TMP;} void insert (trie * root, char * s, int Len) {trie * P = root; For (INT I = 0; I <Len; I ++) {If (p-> next [s [I]-'0'] = NULL) P-> next [s [I]-'0'] = newnode (); P = p-> next [s [I]-'0'];} p-> isword = 1;} void bulid_fail (trie * root) {int head = 0, tail = 0; que [tail ++] = root; root-> fail = NULL; while (Head <tail) {Trie * TMP = que [head ++]; for (INT I = 0; I <10; I ++) {If (TMP-> next [I]) {If (TMP = root) TMP-> next [I]-> fail = root; else {trie * P = TMP-> fail; while (P! = NULL) {If (p-> next [I]) {TMP-> next [I]-> fail = p-> next [I]; break ;} P = p-> fail;} If (P = NULL) TMP-> next [I]-> fail = root ;} if (TMP-> next [I]-> fail-> isword) TMP-> next [I]-> isword = TMP-> next [I]-> fail-> isword; que [tail ++] = TMP-> next [I];} else if (TMP = root) TMP-> next [I] = root; else TMP-> next [I] = TMP-> fail-> next [I] ;}}char STR [100]; ll l, R; ll DP [20] [N] [2] [2]; int bit [25], length; ll DFS (INT Len, int POs, int flag, int Zero, bool limit) {If (LEN <= 0) return flag = 1; if (! Limit & flag & DP [Len] [POS] [zero] [1]! =-1) return DP [Len] [POS] [zero] [1]; If (! Limit &&! Flag & DP [Len] [POS] [zero] [0]! =-1) return DP [Len] [POS] [zero] [0]; ll ans = 0; int up = limit? Bit [Len]: 9; For (INT I = 0; I <= up; I ++) {ans + = DFS (len-1, s [POS]. next [I]-> kind, flag | (s [POS]. next [I]-> isword), 0, limit & (I = up);} If (! Limit) {DP [Len] [POS] [zero] [flag] = ans;} return ans;} ll slove (LL num) {length = 0; while (Num) {bit [++ length] = num % 10; num/= 10;} return DFS (length, 0, 0, 1, true);} ll l, R, K; ll ans; int N; void fuck () {ll mid, TMP = slol-1); While (L <= r) {mid = (L + r)> 1; if (slove (MID)-TMP> = k) {r = mid-1; ans = mid;} else l = Mid + 1 ;}} int main () {// freopen ("1.in"," r ", stdin); While (scanf (" % LLD % d ", & L, & R, & K, & N )! = EOF) {idx = 0; MEM (DP,-1); trie * root = newnode (); For (INT I = 0; I <n; I ++) {scanf ("% s", STR); insert (root, STR, strlen (STR);} bu1__fail (Root); ans =-1; Fuck (); if (ANS =-1) printf ("No such number \ n"); else printf ("% LLD \ n", ANS) ;}return 0 ;}