学步园

思路：二分+数位DP，二分没什么好说的，主要是数位DP，我一开始状态是设计成dp[N][100]，表示i位，后两位为j，这样是会T的。。而其实第二维，只要开3就够了，表示末尾连续的6的个数，因为超过3个就不用转移了

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long ll;

const int N = 15;

int t;
ll n;
int bit[N], bn;

void get(ll x) {
	bn = 0;
	while (x) {
		bit[bn++] = x % 10;
		x /= 10;
	}
	for (int i = 0; i < bn / 2; i++)
		swap(bit[i], bit[bn - i - 1]);
}

ll dp[N][4];

ll solve(ll x) {
	get(x);
	memset(dp, 0, sizeof(dp));
	int pre = 0, flag = 0;
	for (int i = 0; i < bn; i++) {
		for (int j = 0; j < 3; j++) {
			for (int x = 0; x < 10; x++) {
				if (x == 6) dp[i + 1][j + 1] += dp[i][j];
				else dp[i + 1][0] += dp[i][j];
			}
		}
		if (flag) continue;
		for (int x = 0; x < bit[i]; x++) {
			if (x == 6) dp[i + 1][pre + 1]++;
			else dp[i + 1][0]++;
		}
		if (pre == 2 && bit[i] == 6) flag = 1;
		if (bit[i] == 6) pre++;
		else pre = 0;
	}
	ll ans = x + 1;
	for (int j = 0; j < 3; j++)
		ans -= dp[bn][j];
	ans -= !flag;
	return ans;
}

int main() {
	scanf("%d", &t);
	while (t--) {
		scanf("%lld", &n);
		ll l = 666LL, r = 66650000000LL;
		while (l < r) {
			ll mid = (l + r) / 2;
			ll tmp = solve(mid);
			if (solve(mid) >= n) r = mid;
			else l = mid + 1;
		}
		printf("%lld\n", l);
	}
	return 0;
}