Submission #1111427

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Source Code Expand

#include <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>
#include <climits>
#include <ctime>
#include <cassert>
using namespace std;

#define rep(i,a,n) for(int (i)=(a); (i)<(n); (i)++)
#define repq(i,a,n) for(int (i)=(a); (i)<=(n); (i)++)
#define repr(i,a,n) for(int (i)=(a); (i)>=(n); (i)--)
#define all(v) begin(v), end(v)
#define pb(a) push_back(a)
#define fr first
#define sc second
#define INF 2000000000
#define int long long int

#define X real()
#define Y imag()
#define EPS (1e-10)
#define EQ(a,b) (abs((a) - (b)) < EPS)
#define EQV(a,b) ( EQ((a).X, (b).X) && EQ((a).Y, (b).Y) )
#define LE(n, m) ((n) < (m) + EPS)
#define LEQ(n, m) ((n) <= (m) + EPS)
#define GE(n, m) ((n) + EPS > (m))
#define GEQ(n, m) ((n) + EPS >= (m))

typedef vector<int> VI;
typedef vector<VI> MAT;
typedef pair<int, int> pii;
typedef long long ll;

typedef complex<double> P;
typedef pair<P, P> L;
typedef pair<P, double> C;

int dy[]={0, 0, 1, -1};
int dx[]={1, -1, 0, 0};
int const MOD = 1000000007;

namespace std {
    bool operator<(const P& a, const P& b) {
        return a.X != b.X ? a.X < b.X : a.Y < b.Y;
    }
}

int tbl[] = {1, 2, 3, 4, 4, 5, 6, 7, 8, 8};

/*
int calc(string x) {
    if(x == "-1") return 0;
    int ret = 0;
    reverse(x.begin(), x.end());
    bool check[20] = {};
    rep(i,0,x.length()) {
        if(x[i] == '4' || x[i] == '9') {
            rep(j,0,i) check[j] = true;
        }
    }
    rep(i,0,x.length()) {
        int d = x[i] - '0';
        if(check[i]) continue;
        if(i == 0) {
            ret += tbl[d];
            // printf("debug(0): %lld\n", tbl[d]);
        }
        else {
            int temp = 1;
            temp *= tbl[d-1];
            rep(j,0,i) temp *= 8;
            ret += temp;
            // printf("debug(%lld): %lld\n", i, temp);
        }
    }
    return ret;
}

int solve(int a, int b) {
    string s1 = to_string(a-1), s2 = to_string(b);
    int s = b - a + 1;
    int t = calc(s2) - calc(s1);
    // printf("s = %lld, t = %lld\n", s, t);
    return s - t;
}
*/

int solve(int a, int b) {
    string s1 = to_string(a-1), s2 = to_string(b);

    // dp[i][j] = i 桁目まで確定、元の数字未満であることが確定しているかどうか (j で F / T)
    int dp1[20][2] = {}, dp2[20][2] = {};
    dp1[0][0] = 1, dp2[0][0] = 1;
    rep(i,0,s1.length()) {
        int d = s1[i] - '0';
        dp1[i+1][0] = dp1[i][0] * ((d == 4 || d == 9) ? 0 : 1);
        dp1[i+1][1] = dp1[i][0] * tbl[d-1] + dp1[i][1] * tbl[9];
    }

    rep(i,0,s2.length()) {
        int d = s2[i] - '0';
        dp2[i+1][0] = dp2[i][0] * ((d == 4 || d == 9) ? 0 : 1);
        dp2[i+1][1] = dp2[i][0] * tbl[d-1] + dp2[i][1] * tbl[9];
    }

    int s = b - a + 1;
    int t = 0;
    rep(i,0,2) {
        t -= dp1[s1.length()][i];
        t += dp2[s2.length()][i];
    }
    return s - t;
}

signed main() {
    int a, b; cin >> a >> b;
    int ans = solve(a, b);
    cout << ans << endl;
    return 0;
}

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