数位DP CF 55D Beautiful numbers

#include <bits/stdc++.h>

typedef long long ll;
ll dp[20][2520][50];
int digit[20];
int id[2521];
int tot;

int GCD(int a, int b) {
    return b ? GCD (b, a % b) : a;
}
int LCM(int a, int b) {
    return a / GCD (a, b) * b;
}

void init_LCM() {
    tot = 0;
    for (int i=1; i<=2520; ++i) {
        if (2520 % i == 0) {
            id[i] = ++tot;
        }
    }
}

ll DFS(int pos, int val, int lcm, bool limit) {
    if (pos == -1) {
        return val % lcm == 0;
    }
    ll &now = dp[pos][val][id[lcm]];
    if (!limit && now != -1) {
        return now;
    }
    int d = limit ? digit[pos] : 9;
    ll ret = 0;
    for (int i=0; i<=d; ++i) {
        int tval = (val * 10 + i) % 2520;
        int tlcm = i ? lcm / GCD (lcm, i) * i : lcm;
        ret += DFS (pos - 1, tval, tlcm, limit && i == d);
    }
    if (!limit) {
        now = ret;
    }
    return ret;
}

ll solve(ll x) {
    int len = 0;
    if (x == 0) {
        digit[len++] = 0;
    } else {
        while (x) {
            digit[len++] = x % 10;
            x /= 10;
        }
    }
    return DFS (len - 1, 0, 1, true);
}

int main() {
    init_LCM ();
    memset (dp, -1, sizeof (dp));
    int T; scanf ("%d", &T);
    while (T--) {
        ll l, r; std::cin >> l >> r;
        std::cout << solve (r) - solve (l - 1) << ‘\n‘;
    }
    return 0;
}