题面
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\n", ans);
return 0;
}