数位dp。
定义f[i][j][k]表示i位数,各数位的最小公倍数为j,数字%2520的数为k。这样可以进行dfs求解。不过j取为2520会MLE,需要进行离散化。
代码:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define __ ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
const int maxn = 2e5 + 10;
const int maxm = 1e4 + 10;
const ll mod = 998244353;
ll l, r, k, a[20], f[20][50][2530];
int num[200];
map<int, int> mp;
ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); }
ll LCM(ll a, ll b) {
if (min(a, b) == 0)return max(a, b);
return a / gcd(a, b) * b;
}
set<int> s;
ll dp(int p, int l, int m, bool e) {
if (p == 0)return m % num[l] == 0;
if (!e && f[p][l][m] != -1)return f[p][l][m];
ll ans = 0;
for (int i = 0; i <= (e ? a[p] : 9); ++i) {
int L = mp[LCM(num[l], i)], M = (m * 10 + i) % 2520;
ans += dp(p - 1, L, M, e && (i == a[p]));
}
if (!e)f[p][l][m] = ans;
return ans;
}
ll solve(ll x) {
memset(a, 0, sizeof a);
int len = 0;
while (x) {
a[++len] = x % 10;
x /= 10;
}
return dp(len, 1, 0, 1);
}
int main() {
__;
for (int pos = 1; pos < (1 << 10); ++pos) {
ll t = 1;
for (int i = 1; i < 10; ++i) {
if ((1 << i) & pos) t = LCM(t, i);
}
s.insert(t);
}
int cnt = 0;
for (auto i:s) {
num[++cnt] = i;
mp[num[cnt]] = cnt;
}
memset(f, -1, sizeof f);
int _;
cin >> _;
for (int sce = 1; sce <= _; ++sce) {
cin >> l >> r;
cout << solve(r) - solve(l - 1) << endl;
}
return 0;
}