CSES Writing Numbers
// compile: make data
// run: ./data < data.in
// run: ./data
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#ifdef LOCAL
#include <debug/cpglib/print.h>
#define debug(x...) _debug_print(0, #x, x);
#define Debug(x...) _debug_print(1, #x, x);
#define DEBUG(x...) _debug_print(2, #x, x);
std::ifstream terminal("/dev/tty");
#define PP cerr<<"\033[1;30mpause...\e[0m",terminal.ignore();
#else
#define debug(x...)
#define Debug(x...)
#define DEBUG(x...)
#define PP
#endif
template<typename...Args> void print_(Args...args){((cout<<args<<" "),...)<<endl;}
#define rep(i,a,b) for(int i=(a);i<(int)(b);++i)
#define sz(v) ((int)(v).size())
#define print(...) print_(__VA_ARGS__);
#define FIND(a, x) ((find(a.begin(),a.end(),(x))!=a.end())?1:0)
#define cmin(x,...) x=min({(x),__VA_ARGS__})
#define cmax(x,...) x=max({(x),__VA_ARGS__})
#define INTMAX (int)(9223372036854775807)
#define INF (int)(1152921504606846976)
#define NaN (int)(0x8b88e1d0595d51d1)
#define double long double
#define int long long
#define uint unsigned long long
#define endl "\n"
#define MAXN 200010

uint pow10[18];

vector<int> brute(int n) {
    vector<int> cnt(10, 0);
    auto calc = [&](int x) {
        for (; x; x /= 10) ++cnt[x % 10];
    };
    rep(i, 1, n+1) calc(i);
    return cnt;
}

uint solve(uint n) {
    vector<uint> num, cnt(10, 0);
    for (; n; n /= 10) num.emplace_back(n % 10);
    rep(i, 1, num[0]+1) cnt[i] = 1;
    uint cur = num[0];
    rep(i, 1, sz(num)) {
        rep(j, 1, 10) cnt[j] += num[i] * pow10[i-1] * i;
        rep(j, 1, num[i]) cnt[j] += pow10[i];
        cnt[num[i]] += cur + 1;
        cur += num[i] * pow10[i];
    }
    return *max_element(cnt.begin(), cnt.end());
}

int32_t main() {
    ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);

    pow10[0] = 1;
    rep(i, 1, 18) pow10[i] = pow10[i-1] * 10;
    uint n; cin >> n;
    uint l = 1, r = 1e18, ans = 0;
    while (l <= r) {
        uint mid = l + ((r - l) / 2);
        if (solve(mid) <= n) ans = mid, l = mid + 1;
        else r = mid - 1;
    }
    cout << ans << endl;

    return 0;
}

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