#include "koala.h"
#include <bits/stdc++.h>
#pragma GCC Optimize("O3")
#pragma GCC Optimize("unroll-loops")
using namespace std;
int B[100], R[100];
/*
HELPER FUNCTIONS
*/
bool compare(int a, int b, int N, int W) {
fill(B, B + N, 0);
B[a] = B[b] = W;
playRound(B, R);
return (R[b] > W);
}
vector<int> mergesort(vector<int> v, int N, int W) {
if (v.size() == 1) return v;
vector<int> a, b;
a.insert(a.begin(), v.begin(), v.begin() + (v.size() + 1) / 2);
b.insert(b.begin(), v.begin() + (v.size() + 1) / 2, v.end());
a = mergesort(a, N, W), b = mergesort(b, N, W);
vector<int> sorted;
int aptr = 0, bptr = 0;
while (aptr < a.size() && bptr < b.size()) {
if (compare(a[aptr], b[bptr], N, W)) sorted.push_back(a[aptr++]);
else sorted.push_back(b[bptr++]);
}
sorted.insert(sorted.end(), a.begin() + aptr, a.end());
sorted.insert(sorted.end(), b.begin() + bptr, b.end());
return sorted;
}
void split(vector<int> v, int N, int W, int* P, int l = 1, int r = 100) {
if (l == r) P[v[0]] = l;
else {
int x = min((int)sqrt(2 * l), W / (r - l + 1));
fill(B, B + N, 0);
for (int i : v) B[i] = x;
playRound(B, R);
vector<int> less, greater;
for (int i : v) if (R[i] > x) greater.push_back(i);
else less.push_back(i);
split(less, N, W, P, l, l + less.size() - 1);
split(greater, N, W, P, r - greater.size() + 1, r);
}
}
/*
BEGIN ACTUAL FUNCTIONS
*/
/*
Assign a single stone to the first cup.
If Koala picks that cup, it is not the smallest and some other cup has 0.
Otherwise, cup 0 is the smallest cup.
Uses 1 turn
*/
int minValue(int N, int W) {
fill(B, B + N, 0);
B[0] = 1;
playRound(B, R);
if (R[0] < 2) return 0;
else for (int i = 1; i < N; i++) if (!R[i]) return i;
}
/*
Binary search for the max value.
Uses like 4 turns
*/
int maxValue(int N, int W) {
vector<int> v;
for (int i = 0; i < N; i++) v.push_back(i);
while (v.size() != 1) {
int k = W / v.size();
fill(B, B + N, 0);
for (int i : v) B[i] = k;
playRound(B, R);
v.clear();
for (int i = 0; i < N; i++) if (R[i] > k) v.push_back(i);
}
return v[0];
}
/*
Assign positions 0 and 1 x stones until Koala treats them differently.
Binary search for x:
If both 0 and 1 get > x stones, increase x.
If both 0 and 1 get < x stones, decrease x.
Notice how x <= 9 because sum(9..17) > 100.
Uses like 3 turns
*/
int greaterValue(int N, int W) {
int l = 1, r = 9;
while (l != r) {
int mid = (l + r) / 2;
B[0] = B[1] = mid;
playRound(B, R);
if (R[0] > mid && R[1] > mid) l = mid + 1;
else if (R[0] <= mid && R[1] <= mid) r = mid - 1;
else return (R[0] < R[1]);
}
B[0] = B[1] = l;
playRound(B, R);
return (R[0] < R[1]);
}
/*
Subtask 4:
Assign indices i and j 100 stones each to check whether P[i] > P[j].
Use merge sort to get O(Nlog(N)) <= 700 turns
Subtask 5:
We have a recursive strategy that solves a known range [L..R] in
exactly L + R - 1 moves by splitting [L..R] into [L..k] and [k+1..R]
for some k in exactly 1 move.
We do this by assigning each position in the range [L..R]
x = min(sqrt(2 * L), M / (R - L + 1)) stones and then checking which
positions Koala has placed > x stones next to after the round.
We can easily get k this way and we always have 0 < k < L - R + 1
Uses exactly 99 turns
*/
void allValues(int N, int W, int *P) {
if (W == 2 * N) {
vector<int> v;
for (int i = 0; i < N; i++) v.push_back(i);
vector<int> sorted = mergesort(v, N, W / 2);
for (int i = 0; i < N; i++) P[sorted[i]] = i + 1;
} else {
vector<int> v;
for (int i = 0; i < N; i++) v.push_back(i);
split(v, N, W, P);
}
}
