思路：

數位dp能夠求出0到m區間內的符合某個條件的數字個數，在數位dp的條件下通過二分找出符合題意的那個數字

需要注意的是，題目的n雖然是int範圍內的，但是n代表的是符合題意數字的個數，所有實際需要找的範圍應該是大於int的，需要用longlong

#include <iostream>
#include <string>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>

using namespace std;

typedef long long ll;

ll DIG[20], dp[15][15][5];

ll dfs (int cur, int last, int status, bool limit) {
    if (cur < 1) return status;
    if (!limit && dp[cur][last][status] != -1)
        return dp[cur][last][status];

    int End = limit ? DIG[cur] : 9;
    ll ret = 0;

    for (int i = 0; i <= End; ++i)
        ret += dfs (cur-1, i, status && (i != 4 && i != 7), limit && (i == End)); 

    if (!limit)
        dp[cur][last][status] = ret;
    return ret;
}

ll solve (ll x) {
    int m = 0;
    while (x) {
        DIG[++m] = x % 10;
        x /= 10;
    }
    return dfs (m, -1, 1, true)-1; 
}

int main (void)
{
	int t, n;
	scanf ("%d", &t);
	while (t--) {
        scanf ("%d", &n);
        memset (dp, -1, sizeof (dp));
        ll ans = 0, L = 0, R = 10000000000, mid;
        while (L < R) {
            mid = L+(R-L)/2;
            if (solve (mid) >= n) R = mid;  
            else L = mid+1;
        }
        printf ("%lld\n", R);
	}
	return 0;
 }