一道数位dp的题

无链接，朋友问的。

题目

定义“好数”为倍数为7且数字中不含4的数，编写程序在1秒内找出第n（1<n<1e12）个好数

代码

#include<bits/stdc++.h>
#define ll long long

using namespace std;
const int N = 20;
int a[N];
ll f[N][10], n; //f[i][j]:长度为i,余数为j

ll dfs(int pos,int limit, int mod){
	if(!pos) {
        if (mod)    return 0;
        else    return 1;
    }
	if(!limit && f[pos][mod] != -1)    return f[pos][mod];
	ll res = 0, up = limit ? a[pos] : 9;
	for(int i = 0; i <= up;i++) {
		if(i == 4)  continue;	
		res += dfs(pos - 1,limit && i == up, (mod * 10 + i) % 7);		
	}
	return limit ? res : f[pos][mod] = res;
}

ll dp(ll n) {
	memset(f, -1, sizeof f);
	int len = 0;
	while(n)    a[++len] = n % 10, n /= 10;
	return dfs(len ,1, 0) - 1;
}

void solve () {
    cin >> n;
	ll l = 1, r = 1e18;
	while (l < r) {
		ll mid = (l + r) >> 1;
		if(dp(mid) >= n)    r = mid;
		else    l = mid + 1;
	}
	//cout << n << ' ' << l << "\n";
    cout << l << endl;
}

int main() {
	//for (int i = 1; i <= 100; i++)  n = i, solve ();
    solve ();
	return 0;
}