#include "Encoder.h"
#include <bits/stdc++.h>
using namespace std;
void Encode(int N, int A[], int B[]) {
// Euler tour, send information about the left end
// and the interval length. This scores 30 points.
// To improve, notice that representing the interval
// length is wasteful - we can add dummy nodes to
// force interval lengths into a smaller value range.
// If we choose to represent the interval lengths by
// r^0, r^1, r^2, r^3, ..., then the root node's size
// can be multiplied by a factor of r^d, where d is
// the depth of the subtree (d <= 18). Thus for the
// enter time, we have N r^d values, and for the interval
// length, we have log(N r^d, r) values - thus the maximum
// value is N r^d log(N r^d, r) = N r^d (log(N, r) + d).
// Optimizing this function for N = 250000 and d = 18 yields
// r = 1.053 as the optimal ratio, and we need 27.2877 bits.
// Accounting for integer rounding, this is below the needed
// 28 bits for perfect score.
static vector<int> lengths;
auto GenerateLengths = [&]() {
if (!lengths.empty()) {
return;
}
const auto OptimalRatio = [&]() {
const auto Eval = [&](double x) -> double {
const double MAXN = 250000;
const double MAXD = 18;
return MAXN * pow(x, MAXD) * (log(MAXN) / log(x) + MAXD);
};
double lo = 1, hi = 2;
for (int rep = 0; rep < 200; rep++) {
double md1 = (lo + lo + hi) / 3;
double md2 = (lo + hi + hi) / 3;
if (Eval(md1) < Eval(md2)) {
hi = md2;
} else {
lo = md1;
}
}
return lo;
};
const int MAXN = 250000;
const int MAXD = 18;
const double ratio = OptimalRatio();
int cur = 1;
int greater_than_MAXN = 0;
while (greater_than_MAXN < MAXD) {
if (cur > MAXN) {
greater_than_MAXN += 1;
}
lengths.emplace_back(cur);
cur = max(cur + 1,(int) round(cur * ratio));
}
};
vector<vector<int>> adj(N);
for (int i = 0; i + 1 < N; i++) {
adj[A[i]].emplace_back(B[i]);
adj[B[i]].emplace_back(A[i]);
}
int timer = 0;
vector<int> st(N), etlen(N);
function<void(int, int)> Dfs = [&](int u, int p) {
st[u] = timer++;
for (auto v : adj[u]) if (v != p) {
Dfs(v, u);
}
etlen[u] = lower_bound(begin(lengths), end(lengths), timer - st[u]) - begin(lengths);
timer = st[u] + lengths[etlen[u]];
};
GenerateLengths();
Dfs(0, -1);
for (int i = 0; i < N; i++) {
Code(i, st[i] * lengths.size() + etlen[i]);
}
}
#include "Device.h"
#include <bits/stdc++.h>
using namespace std;
void InitDevice() {}
int Answer(long long S, long long T) {
// Euler tour, send information about the left end
// and the interval length. This scores 30 points.
// To improve, notice that representing the interval
// length is wasteful - we can add dummy nodes to
// force interval lengths into a smaller value range.
// If we choose to represent the interval lengths by
// r^0, r^1, r^2, r^3, ..., then the root node's size
// can be multiplied by a factor of r^d, where d is
// the depth of the subtree (d <= 18). Thus for the
// enter time, we have N r^d values, and for the interval
// length, we have log(N r^d, r) values - thus the maximum
// value is N r^d log(N r^d, r) = N r^d (log(N, r) + d).
// Optimizing this function for N = 250000 and d = 18 yields
// r = 1.053 as the optimal ratio, and we need 27.2877 bits.
// Accounting for integer rounding, this is below the needed
// 28 bits for perfect score.
static vector<int> lengths;
auto GenerateLengths = [&]() {
if (!lengths.empty()) {
return;
}
const auto OptimalRatio = [&]() {
const auto Eval = [&](double x) -> double {
const double MAXN = 250000;
const double MAXD = 18;
return MAXN * pow(x, MAXD) * (log(MAXN) / log(x) + MAXD);
};
double lo = 1, hi = 2;
for (int rep = 0; rep < 200; rep++) {
double md1 = (lo + lo + hi) / 3;
double md2 = (lo + hi + hi) / 3;
if (Eval(md1) < Eval(md2)) {
hi = md2;
} else {
lo = md1;
}
}
return lo;
};
const int MAXN = 250000;
const int MAXD = 18;
const double ratio = OptimalRatio();
int cur = 1;
int greater_than_MAXN = 0;
while (greater_than_MAXN < MAXD) {
if (cur > MAXN) {
greater_than_MAXN += 1;
}
lengths.emplace_back(cur);
cur = max(cur + 1,(int) round(cur * ratio));
}
};
GenerateLengths();
int stx, etlenx, sty, etleny;
stx = S / lengths.size();
etlenx = lengths[S % lengths.size()];
sty = T / lengths.size();
etleny = lengths[T % lengths.size()];
if (sty <= stx && stx + etlenx <= sty + etleny) {
return 0;
} else if (stx <= sty && sty + etleny <= stx + etlenx) {
return 1;
} else {
return 2;
}
}
