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Sol

在AC自動機上跑數位DP
設\(f[i][j][0/1]\)表示到\(n的第i位\)當前匹配到\(AC自動機的j節點\)的方案
轉移就在AC自動機上跑
註意不能有前導零，可能有這種情況\(000000\)不能存在那麽前導零就有問題
所以要單獨把小於\(n\)的位數的數單獨算出來，等於\(n\)的位數的數單獨算出來最後加起來

# 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;
}

Bzoj3530: [Sdoi2014]數數