[ZOJ4317] BCD Code [数位dp] [AC自动机]

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没想到我这个老年选手还能够一发敲对（其实是两发，第一次忘记在最后减的时候取模了。。
很容易想到是数位dp，一边dfs一边在ac自动机那个trie上面跑就行了，没有思维难度
当然也可以预处理不过我觉得这么写好像更顺手（

problemId有两个，一个是3494，一个是4317，我就信链接上面那个吧（

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<queue>
#include<cmath>
#include<ctime>
using namespace std;
const int mod = 1000000009;
int T, n, len, tot;
char A[205],B[205];
int bit[205];
char s[25];
bool code[2005][25];
int fail[2005];
int nxt[2005][2];
bool ed[2005];
inline void insert(int pos, const int& id, int step) {
	if(step>len) { ed[pos]=1; return;}
	insert(nxt[pos][code[id][step]] ? nxt[pos][code[id][step]] : (nxt[pos][code[id][step]] = ++tot), id, step + 1);
}
queue<int>Q;
void directfail() {
	if(nxt[0][0])Q.push(nxt[0][0]);
	if(nxt[0][1])Q.push(nxt[0][1]);
	while(!Q.empty()) {
		int x = Q.front(); Q.pop(); ed[x]|=ed[fail[x]];
		for(int i = 0; i <= 1; ++i) {
			if(nxt[x][i])fail[nxt[x][i]]=nxt[fail[x]][i], Q.push(nxt[x][i]);
			else nxt[x][i]=nxt[fail[x]][i];
		}
	}
}
int F[205][2005];
int dfs(int step, int pos, bool lead, bool limit) {
	if (!step) return 1;
	if (!lead && !limit && F[step][pos] != -1) return F[step][pos];
	int ret = 0, up = limit ? bit[step] : 9;
	for (int p, i = 0; i <= up; ++i) {
		
		p = pos;
		
		if (!(lead&&!i)) {
			for(int j = 3; j >= 0; --j) {
				p = nxt[p][(i>>j)&1];
				if (ed[p]) break;
			}
			if (ed[p]) continue;
		}
		
		ret += dfs(step - 1, p, lead && i == 0, limit && i == up);
		ret %= mod;
	}
	if(!lead && !limit) F[step][pos] = ret;
	return ret;
}
int solve(char *x) {
	bit[0] = strlen(x+1);
	if(bit[0]==1&&x[1]=='0') return 1;
	for(int i = 1; i <= bit[0]; ++i) bit[i] = x[i] - 48;
	reverse(bit+1,bit+1+bit[0]);
	return dfs(bit[0], 0, true, true);
}
char *sub(char *x) {
	int t = strlen(x+1);
	--x[t];
	for(int i = t; i >= 1; --i) {
		if(x[i] < '0') x[i] += 10, --x[i-1];
		else break;
	}
	int p = 1;
	while(!x[p] && p<t)++p;
	return x+p-1;
}
int main() {
	scanf("%d", &T);
	while (T--) {
		memset(nxt, 0, sizeof(nxt));
		memset(fail, 0, sizeof(fail));
		memset(ed, 0, sizeof(ed));
		memset(F, -1, sizeof(F));
		memset(A, 0, sizeof(A));
		memset(B, 0, sizeof(B));
		tot = 0;
		scanf("%d", &n);
		for (int i = 1; i <= n; ++i) {
			memset(s,0,sizeof(s));
			scanf(" %s", s+1);
			len = strlen(s+1);
			for(int j = 1; j <= len; ++j) code[i][j] = s[j] - '0';
			insert(0, i, 1);
		}
		directfail();
		scanf(" %s %s", A+1, B+1);
		printf("%d\n", ( solve(B) - solve(sub(A)) + mod ) % mod ); //**
	}
	return 0;
}