題解來源於rgnoH：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define ll long long
#define rep(i,x,y) for(ll i=(x);i<=(y);i++)
#define repl(i,x,y) for(ll i=(x);i<(y);i++)
#define repd(i,x,y) for(ll i=(x);i>=(y);i--)
using namespace std;

const ll N=25;
const ll M=15;
const ll Inf=1e18;

ll n,w[N],f[N][M][M][2][2][2][2];

inline ll read() {
	ll x=0;char ch=getchar();bool f=0;
	while(ch>'9'||ch<'0'){if(ch=='-')f=1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
	return f?-x:x;
}

void split(ll x) {
	w[0]=0;
	while(x) {
		w[++w[0]]=x%10;x/=10;
	}
	rep(i,1,w[0]/2) swap(w[i],w[w[0]+1-i]);
}

ll dp(ll len,ll sec,ll thi,ll lim,ll lim1,ll lim2,ll lim3) {
	if(~f[len][sec][thi][lim][lim1][lim2][lim3]) return f[len][sec][thi][lim][lim1][lim2][lim3];
	
	if(len==w[0]+1) {
		if(lim1&&(!(lim2&lim3))) return true;
		return false;
	}
	
	ll ans=0,max_num=lim?w[len]:9;
	
	ll nlim,nlim1,nlim2,nlim3;
	
	rep(i,0,max_num) {
		if(!lim1) {
			if(lim&&i==w[len]) nlim=1; else nlim=0;
			if(i) nlim1=1; else nlim1=0; nlim2=nlim3=0;
			
			ans+=dp(len+1,i,sec,nlim,nlim1,nlim2,nlim3);
		} else {
			if(lim&&i==w[len]) nlim=1; else nlim=0;
			if(i) nlim1=1; else nlim1=lim1;
			if(i==sec) nlim2=1; else nlim2=lim2;
			if(i==sec-1) nlim3=1; else nlim3=lim3;
			if(!(i==3&&sec==3&&thi==2)&&(!(nlim2&nlim3))) ans+=dp(len+1,i,sec,nlim,nlim1,nlim2,nlim3);
		}
	}
	
	return f[len][sec][thi][lim][lim1][lim2][lim3]=ans;
}

bool check(ll x) {
	split(x);
	memset(f,-1,sizeof(f));
	return dp(1,11,11,1,0,0,0)>=n;
}

int main() {
	n=read();
	
	ll l=1,r=Inf;
	while(l<=r) {
		ll mid=l+r>>1;
		if(check(mid)) r=mid-1;
		else l=mid+1;
	}
	
	printf("%lld",l);

	return 0;
}