Rational Approximation.cpp

#include<bits/stdc++.h>
using namespace std;

using ll = long long;
/**
Given n and a real number x >= 0, returns the closest rational approximation p/q s.t. p, q <= n.
It will obey that |p/q - x| is minimum for p, q <= n
Time: O(log n)
**/
using ld = long double;
pair<ll, ll> approximate(ld x, ll n) {
  ll LP = 0, LQ = 1, P = 1, Q = 0, inf = LLONG_MAX; ld y = x;
  while (1) {
    ll lim = min(P ? (n - LP) / P : inf, Q ? (n - LQ) / Q : inf),
    a = (ll)floor(y), b = min(a, lim),
    NP = b * P + LP, NQ = b * Q + LQ;
    if (a > b) {
      // If b > a/2, we have a semi-convergent that gives us a
      // better approximation; if b = a/2, we *may* have one.
      // Return {P, Q} here for a more canonical approximation.
      return (abs(x - (ld)NP / (ld)NQ) < abs(x - (ld)P / (ld)Q)) ?
        make_pair(NP, NQ) : make_pair(P, Q);
    }
    if (abs(y = 1 / (y - (ld)a)) > 3 * n) {
      return {NP, NQ};
    }
    LP = P; P = NP;
    LQ = Q; Q = NQ;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t; cin >> t;
  while (t--) {
    long double x; cin >> x;
    ll n = 1e9;
    auto ans = approximate(x, n);
    cout << ans.first << ' ' << ans.second << '\n';
  }
  return 0;
}
// https://official.contest.yandex.ru/opencupXVIII/contest/5457/problems/E?lang=en