Bzoj3530: [Sdoi2014]数数

题面

传送门

Sol

在AC自动机上跑数位DP
设(f[i][j][0/1])表示到(n的第i位)当前匹配到(AC自动机的j节点)的方案
转移就在AC自动机上跑
注意不能有前导零，可能有这种情况(000000)不能存在那么前导零就有问题
所以要单独把小于(n)的位数的数单独算出来，等于(n)的位数的数单独算出来最后加起来

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int Zsy(1e9 + 7), _(1510);

int n, m, ch[10][_], tot, fail[_], f[3][_][_], ans;
bool cnt[_];
char s[_], T[_];
queue <int> Q;

IL void Insert(){
	RG int x = 0, len = strlen(s);
	for(RG int i = 0; i < len; ++i){
		if(!ch[s[i] - '0'][x]) ch[s[i] - '0'][x] = ++tot;
		x = ch[s[i] - '0'][x];
	}
	cnt[x] = 1;
}

IL void GetFail(){
	for(RG int i = 0; i <= 9; ++i) if(ch[i][0]) Q.push(ch[i][0]);
	while(!Q.empty()){
		RG int fa = Q.front(); Q.pop();
		for(RG int i = 0; i <= 9; ++i)
			if(ch[i][fa]) fail[ch[i][fa]] = ch[i][fail[fa]], Q.push(ch[i][fa]);
			else ch[i][fa] = ch[i][fail[fa]];
		cnt[fa] |= cnt[fail[fa]];
	}
}

IL void Calc(){
	f[1][0][0] = f[2][0][0] = 1;
	for(RG int i = 0; i < n; ++i)
		for(RG int j = 0; j <= tot; ++j){
			if(cnt[j]) continue;
			for(RG int l = 0; l <= 9; ++l){
				RG int p = ch[l][j];
				if(!(i + l) || cnt[p]) continue;
				(f[2][i + 1][p] += f[2][i][j]) %= Zsy;
			}
		}
	for(RG int i = 1; i < n; ++i)
		for(RG int j = 0; j <= tot; ++j)
			(ans += f[2][i][j]) %= Zsy;
	for(RG int i = 0; i < n; ++i)
		for(RG int j = 0; j <= tot; ++j){
			if(cnt[j]) continue;
			for(RG int l = 0; l <= 9; ++l){
				RG int p = ch[l][j];
				if(!(i + l) || cnt[p]) continue;
				(f[0][i + 1][p] += f[0][i][j]) %= Zsy;
				if(l + '0' == T[i + 1]) (f[1][i + 1][p] += f[1][i][j] % Zsy) %= Zsy;
				if(l + '0' < T[i + 1]) (f[0][i + 1][p] += f[1][i][j]) %= Zsy;
			}
		}
}

int main(RG int argc, RG char* argv[]){
	scanf(" %s%d", T + 1, &m); n = strlen(T + 1);
	for(RG int i = 1; i <= m; ++i) scanf(" %s", s), Insert();
	GetFail(); Calc();
	for(RG int i = 0; i <= tot; ++i) (ans += (f[0][n][i] + f[1][n][i]) % Zsy) %= Zsy;
	printf("%d
", ans);
	return 0;
}