Codeforces 833 C

思路：

首先，inedible tails 的个数最多为C(18+9，9）个（用隔板法），所以我们暴力出所有的 inedible tails，然后检查一下在[L, R]这段区间是否存在这个inedible tails

检查的时候用了和数位dp差不多的方法，设一个下界和上界，只要之前的既没有达到上界也没有达到下界，后面就可以随便填了，说明存在

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

int l[25], r[25], f[25], ans = 0, cnt = 0;
bool check(int pos, bool Llimit, bool Rlimit) {
    if(!Llimit && !Rlimit) return true;
    if(pos == 0) return true;
    int down = Llimit ? l[pos] : 0;
    int up = Rlimit ? r[pos] : 9;
    for (int i = down; i <= up; i++) {
        if(f[i]) {
            f[i]--;
            bool flag = check(pos-1, Llimit&&i==l[pos], Rlimit&&i==r[pos]);
            f[i]++;
            if(flag) return true;

        }
    }
    return false;
}
void dfs(int pos, int res) {
    if(pos == 9) {
        f[pos] = res;
        if(check(cnt, 1, 1)) ans++;
        return ;
    }
    for (int i = 0; i <= res; i++) {
        f[pos] = i;
        dfs(pos+1, res - i);
    }
}
int solve(LL L, LL R) {
    cnt = 0;
    while(L) {
        l[++cnt] = L%10;
        L /= 10;
    }

    cnt = 0;
    while(R) {
        r[++cnt] = R%10;
        R /= 10;
    }
    ans = 0;
    dfs(0, cnt);
    return ans;
}
int main() {
    LL L, R;
    scanf("%lld %lld", &L, &R);
    printf("%d
", solve(L, R));
    return 0;
}