Description
一個數x各個數位上的數之積記為f(x) <不含前導零>
求[L,R)中滿足0 < f(x) <= n的數的個數

Sample Input
5
19 22

Sample Output
1

突然發現自己沒寫過數位DP的blog，寫一篇！！！
首先我們發現直接開陣列n太大會爆炸。。。
然後我們發現其實1~9的質因數是有限的：2、3、5、7
那你開一個五維陣列f[i][x2][x3][x5][x7]表示到第i位，有x2個2，x3個3，x5個5，x7個7。
然後就數位DP亂搞。。。

#include <cstdio>
 

#include <cstring>

using namespace std;
typedef long long LL;

LL f[20][31][21][13][12];
int n, a[20], len;
int k2, k3, k5, k7;

LL dfs(int x, int x2, int x3, int x5, int x7, bool pd, bool limit) {
    LL oo = 1;
    for(int i = 1; i <= x2; i++) oo *= 2;
    for(int i = 1; i <= x3; i++) oo *= 3
 
;
    for(int i = 1; i <= x5; i++) oo *= 5;
    for(int i = 1; i <= x7; i++) oo *= 7;
    if(oo > n) return 0;
    if(x == 0) {
        if(oo <= n && !pd) return 1;
        else return 0;
    }
    if(f[x][x2][x3][x5][x7] != -1 && !limit && !pd) return f[x][x2][x3][x5][x7];
    int
 
 up = 9; LL ans = 0; if(limit) up = a[x];
    if(pd) ans += dfs(x - 1, x2, x3, x5, x7, 1, limit);
    for(int i = 1; i <= up; i++) {
        bool uu = 0;
        if(limit == 1 && i == up) uu = 1;
        int y2 = x2, y3 = x3, y5 = x5, y7 = x7;
        if(i == 2) y2++; if(i == 3) y3++; if(i == 4) y2 += 2; if(i == 5) y5++; if(i == 6) y2++, y3++;
        if(i == 7) y7++; if(i == 8) y2 += 3; if(i == 9) y3 += 2;
        ans += dfs(x - 1, y2, y3, y5, y7, 0, uu);
    }
    if(!limit && !pd) f[x][x2][x3][x5][x7] = ans;
    return ans;
}

LL solve(LL x) {
    if(x == 0) return 0;
    len = 0;
    while(x) a[++len] = x % 10, x /= 10;
    LL ans = 0;
    memset(f, -1, sizeof(f));
    for(int i = 0; i <= a[len]; i++) {
        bool h1 = 0, h2 = 0;
        if(i == 0) h1 = 1; if(i == a[len]) h2 = 1;
        int y2 = 0, y3 = 0, y5 = 0, y7 = 0;
        if(i == 2) y2++; if(i == 3) y3++; if(i == 4) y2 += 2; if(i == 5) y5++; if(i == 6) y2++, y3++;
        if(i == 7) y7++; if(i == 8) y2 += 3; if(i == 9) y3 += 2;
        ans += dfs(len - 1, y2, y3, y5, y7, h1, h2);
    }
    return ans;
}

int main() {
    scanf("%d", &n);
    LL x, y; scanf("%lld%lld", &x, &y);
    printf("%lld\n", solve(y - 1) - solve(x - 1));
    return 0;
}