1
  2
  3
  4
  5
  6
  7
  8
  9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27
 28
 29
 30
 31
 32
 33
 34
 35
 36
 37
 38
 39
 40
 41
 42
 43
 44
 45
 46
 47
 48
 49
 50
 51
 52
 53
 54
 55
 56
 57
 58
 59
 60
 61
 62
 63
 64
 65
 66
 67
 68
 69
 70
 71
 72
 73
 74
 75
 76
 77
 78
 79
 80
 81
 82
 83
 84
 85
 86
 87
 88
 89
 90
 91
 92
 93
 94
 95
 96
 97
 98
 99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cassert>
#include <iostream>
#include <algorithm>
#include <iterator>
#include <string>
#include <vector>
#include <queue>
#include <bitset>
#include <utility>
#include <stack>
#include <numeric>

using namespace std;
typedef long long LL;
typedef vector<int> VI;

#define FOR(v,p,k) for(int v=(p);v<=(k);++v)
#define FORD(v,p,k) for(int v=(p);v>=(k);--v)
#define REP(i,n) for(int i=0;i<(n);++i)
#define PB push_back

int new_key(int old_key, int digit) {
    switch( digit ) {
        case 0: return old_key;
        case 1: return old_key;
        case 2: return old_key | 1;
        case 3: return old_key | 2;
        case 4: return old_key | 4;
        case 5: return old_key | 8;
        case 6: return old_key | 3;
        case 7: return old_key | 16;
        case 8: return old_key | 32;
        case 9: return old_key | 64;
        default: assert(false);
    }
}

const int max_key = 127;
const VI digits_for_keys = {2, 3, 4, 5, 7, 8, 9};

bool matches_key(int key, int digit) {
    return (new_key(0, digit) & key) > 0;
}

void remove_redundant(VI &digits) {
    if (count(begin(digits), end(digits), 9)) {
        digits.erase(remove(begin(digits), end(digits), 3), end(digits));
    }

    if (count(begin(digits), end(digits), 8)) {
        digits.erase(remove(begin(digits), end(digits), 4), end(digits));
        digits.erase(remove(begin(digits), end(digits), 2), end(digits));
    }

    if (count(begin(digits), end(digits), 4)) {
        digits.erase(remove(begin(digits), end(digits), 2), end(digits));
    }
}

// dp copied from (and changed): https://www.hackerrank.com/topics/digit-dp
const int MODULO = 5*7*8*9;
LL dp[20][2][max_key+1][MODULO][2];//index (of digit in number), smaller (to control if we can use high digits), key (based on used digits set), modulo remainder, used_zero (already, 0 can be used only at the beginning)

void dp_part(const string &s, int index, int smaller, int pow10) {
    if (index == s.length()) {
        int key = 0, m = 0, used_zero = 0;
        dp[index][smaller][key][m][used_zero] = 1;
        return;
    }

    int limit=9;
    if (smaller) {
        limit=s[index]-'0';
    }

    FOR(i, 0, limit) {
        int next_smaller;
        if(i<s[index]-'0') {
            next_smaller=0;
        } else {
            next_smaller=smaller;
        }
        FOR(prev_key, 0, max_key) {
            int key = new_key(prev_key, i);
            REP(prev_mod, MODULO) {
                //for index = s.lenght() - 1 only for prev_mode == 0 we will have non zero entries for index+1
                int m = ((pow10 * i) % MODULO + prev_mod) % MODULO;
                REP(used_zero, 2) {
                    if (i > 0 && used_zero == 0) {
                        dp[index][smaller][key][m][used_zero] += dp[index+1][next_smaller][prev_key][prev_mod][used_zero];
                    }

                    if (i == 0 && used_zero == 1) {
                        dp[index][smaller][key][m][used_zero] += dp[index+1][next_smaller][prev_key][prev_mod][used_zero];
                        dp[index][smaller][key][m][used_zero] += dp[index+1][next_smaller][prev_key][prev_mod][1-used_zero];
                    }
                }
            }
        }
    }
}

void dp_solve(LL n) {
    memset(dp,0,sizeof(dp));
    string s=to_string(n);
    int pow10 = 1;
    FORD(index, s.length(), 0) {
        REP(smaller, 2) {
            dp_part(s, index, smaller, pow10);
        }
        //for index == s.lenght() we don't use pow10
        if (index < s.length()) {
            pow10 = (pow10 * 10) % MODULO;
        }
    }
}

LL solve(LL n) {
    dp_solve(n);
    LL res = 0;
    VI digits_used;
    digits_used.reserve(10);
    FOR(key, 0, max_key) {
        digits_used.clear();
        for(int d: digits_for_keys) {
            if (matches_key(key, d)) {
                digits_used.PB(d);
            }
        }
        remove_redundant(digits_used);
        int multi = accumulate(begin(digits_used), end(digits_used), 1, multiplies<int>());
        REP(i, MODULO/multi) {
            int m = i*multi;
            REP(used_zero, 2) {
                int index = 0, smaller = 1;
                res += dp[index][smaller][key][m][used_zero];
            }
        }
    }
    return res;
}


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