count n length numbers without leading zeros that used all digit(mask dp)

#include <bits/stdc++.h>

using namespace std;
#define ll long long int
int n;
const int mx = 1000;

bool check(int mask) {
    return (bool) (mask & ((1 << 10) - 1));
}

int dp[mx][(1 << 10) + 2];

int solve(int pos, int mask) {
    if (pos >= n) {
        return check(mask);
    }
    if (dp[pos][mask] != -1)return dp[pos][mask];
    int low = 0;
    if (pos == 0) {
        low = 1;
    }
    int res = 0;
    for (int i = low; i < 10; i++) {
        int val = solve(pos + 1, mask | (1 << pos));
        res += val;
    }
    return dp[pos][mask] = res;
}

int main() {
    memset(dp, -1, sizeof(dp));
    cin >> n;
    cout << solve(0, 0);
    return 0;
}


//how many permutation of s<=10(unique digit number)are divisible by d;

#include <bits/stdc++.h>

using namespace std;
#define ll long long int
string s;
int d;
const int mx = 1005;
int dp[(1 << 10) + 2][mx];

int solve(int mask, int mod) {
    int on_bit_cnt = __builtin_popcount(mask);
//    cout << on_bit_cnt << " " << mod << endl;
    if (on_bit_cnt == s.size()) {
        return (mod == 0);
    }
    if (dp[mask][mod] != -1)return dp[mask][mod];
    int res = 0;
    for (int i = 0; i < s.size(); i++) {
        if (mask & (1 << i))continue;
        int val = solve(mask | (1 << i), (mod * 10 + s[i] - '0') % d);
        res += val;
    }
    return dp[mask][mod] = res;
}

int main() {
    cin >> s >> d;
    memset(dp, -1, sizeof(dp));
    cout << solve(0, 0) << endl;
    return 0;
}