139 lines
2.5 KiB
C++
139 lines
2.5 KiB
C++
#include <bits/stdc++.h>
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// https://codeforces.com/blog/entry/96344
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#pragma GCC optimize("O2,unroll-loops")
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#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
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using namespace std;
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template <typename... Args>
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void dbg(std::string const &str, Args &&...args) {
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std::cout << std::vformat(str, std::make_format_args(args...));
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}
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template <typename T>
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void dbg(T const &t) {
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std::cout << t;
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}
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template <std::ranges::range T>
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void dbgln(T const &t) {
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if constexpr (std::is_convertible_v<T, char const *>) {
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std::cout << t << '\n';
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} else {
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for (auto const &e : t) {
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std::cout << e << ' ';
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}
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std::cout << '\n';
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}
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}
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void dbgln() {
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std::cout << '\n';
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}
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template <typename... Args>
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void dbgln(std::string const &str, Args &&...args) {
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dbg(str, std::forward<Args>(args)...);
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cout << '\n';
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}
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template <typename T>
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void dbgln(T const &t) {
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dbg(t);
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cout << '\n';
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}
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template <typename T>
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constexpr T MIN = std::numeric_limits<T>::min();
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template <typename T>
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constexpr T MAX = std::numeric_limits<T>::min();
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template <typename T>
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static T sc(auto &&x) {
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return static_cast<T>(x);
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}
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#define ff first
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#define ss second
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#define eb emplace_back
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#define ll long long
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#define ld long double
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#define vec vector
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#define endl '\n'
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#define all(x) (x).begin(), (x).end()
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#define rall(x) (r).rbegin(), (x).rend()
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#define sz(x) static_cast<int>((x).size())
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#define FOR(a, b, c) for (int(a) = (b); (a) < (c); ++(a))
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#define ROF(a, b, c) for (int(a) = (b); (a) > (c); --(a))
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std::random_device rd;
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std::mt19937 gen(rd());
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long long power(long long base, long long exp) {
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long long result = 1;
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while (exp > 0) {
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if (exp & 1)
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result *= base;
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base *= base;
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exp >>= 1;
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}
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return result;
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}
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ll log_baseb(ll x, ll b) {
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ll res = 0;
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while (x >= b) {
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x /= b;
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res++;
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}
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return res;
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}
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ll ceil_log_baseb(ll x, ll base) {
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ll res = 0, power = 1;
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while (power < x) {
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power *= base;
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res++;
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}
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return res;
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}
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void solve() {
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ll l, r;
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cin >> l >> r;
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ll ans = l == 1 ? 1 : 1 + log_baseb(l, 3);
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if (r - l + 1 > 1)
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ans *= 2;
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// add on count to reduce range [l+1,r] to 0
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ll L = l + 1;
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while (L <= r) {
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ll B = min(r + 1, (ll)power(3, 1 + log_baseb(L, 3)));
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// dbgln("floor(log({})/log(3))={}", L, log_baseb(L, 3));
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// dbgln("L={}, B={}", L, B);
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ans += ceil_log_baseb(B, 3) * (B - L);
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L = B;
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}
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dbgln(ans);
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}
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int main() {
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cin.tie(nullptr)->sync_with_stdio(false);
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int t = 1;
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cin >> t;
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while (t--) {
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solve();
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}
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return 0;
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}
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