HDU-2089-不要62

这题就是数位dp，这里不多解释了，传送门在此：
https://blog.csdn.net/qq_41090676/article/details/86821500

#include <iostream>
#include <cstring>
using namespace std;
typedef long long ll;
int digit[20];
long long dp[20][2];

ll dfs(int len,bool is6,bool limit)
{
    if (len==0)
        return 1;
    if (!limit&&dp[len][is6])
        return dp[len][is6];
    ll cnt = 0, up_bound = (limit ? digit[len] : 9);
    for (int i = 0; i <= up_bound;i++) {
        if (is6&&i==2)
            continue;
        if (i==4)
            continue;
        cnt += dfs(len - 1, i == 6, limit && i == up_bound);
    }
    if (!limit)
        dp[len][is6] = cnt;
    return cnt;
}

ll solve(ll num)
{
    int k = 0;
    while (num) {
        digit[++k] = num % 10;
        num /= 10;
    }
    return dfs(k, false, true);
}

int main()
{
    ll n, m;
    while (cin>>n>>m&&n+m) {
        memset(dp, 0, sizeof(dp));
        cout << solve(m) - solve(n-1) << endl;
    }
    return 0;
}