思路：
数位dp
代码：
#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 LL L, R, L1, R1;
int tot = ;
int a[];
map<LL, LL>dp[];
LL dfs(int pos, LL mul, bool limit, bool zero) {
if(pos == ) {
if(zero) mul = ;
if(L1 <= mul && mul <= R1) return ;
else return ;
}
if(!limit && !zero && dp[pos].find(mul) != dp[pos].end()) return dp[pos][mul];
int up = ;
if(limit) up = a[pos];
LL ans = ;
for (int i = ; i <= up; i++) {
if(zero) {
if(i == ) ans += dfs(pos-, mul, limit && i == up, zero && i == );
else ans += dfs(pos-, mul*i, limit && i == up, zero && i == );
}
else ans += dfs(pos-, mul*i, limit && i == up, zero && i == );
}
if(!limit && !zero) dp[pos][mul] = ans;
return ans;
}
LL solve(LL n) {
tot = ;
if(n == -) return ;
if(n == ) tot++;
while(n) {
a[++tot] = n%;
n /= ;
}
return dfs(tot, , , );
}
int main() {
scanf("%lld %lld %lld %lld", &L, &R, &L1, &R1);
printf("%lld\n", solve(R) - solve(L-));
return ;
}