(Analysis by Benjamin Qi)

Farmer Nhoj's cows partition the number line into $M+1$ intervals. The general idea is to find the answers for each of these intervals independently, and then combine them.

Firstly, note that John can claim the total tastiness of the leftmost interval with a single cow. Similar reasoning holds for the rightmost interval. Every other of the $M-1$ intervals has one of Nhoj's cows at both of its endpoints, so one of John's cows might not be enough to claim all of the tastinesses within it. But placing two cows will definitely be enough (one just to the right of the left endpoint of the interval, and another just to the left of the right endpoint of the interval).

Assume all the patches and cows are sorted in increasing order of position, and consider the interval $(f_{x-1},f_x)$. Probably the trickiest part of the solution involves computing the maximum tastiness John can claim if he places only one cow in this interval. If John places a cow at $q\in (f_{x-1},f_x)$, he claims the tastinesses of all patches with positions in the range $\left(\frac{f_{x-1}+q}{2},\frac{f_{x}+q}{2}\right)$. Note that the length of this range ("window") is equal to $\frac{f_x-f_{x-1}}{2}$ regardless of $q$. We can slide this constant-length window from left to right and keep track of how the sum of tastinesses changes as patches enter and exit this window. The one-cow answer for this interval will equal the maximum sliding window sum that we ever encounter.

Now that we've computed the one-cow and two-cow answers for every interval, it remains to combine these to get the answer to the original problem. If John was restricted to placing at most one cow in every interval, then we could create a list of all the one-cow answers, sort it in decreasing order, and sum its first $N$ elements. It turns out that a slight modification of this strategy works when we allow John to place up to two cows in every interval: add the quantity $(\text{two-cow answer} - \text{one-cow answer})$ for each interval to the list before sorting it (if there are ties, put quantities of the form $\text{one-cow answer}$ before quantities of the form $(\text{two-cow answer} - \text{one-cow answer})$).

The reason these works is because $2\cdot (\text{one-cow answer})\ge (\text{two-cow answer})$ for any interval; in other words, adding a second cow to an interval can't increase the total tastiness more than adding the first cow did. This implies that if a prefix of the list contains the quantity $(\text{two-cow answer} - \text{one-cow answer})$ for an interval, it will also contain the quantity $\text{one-cow answer}$ for that same interval.

My code (with a single vector for both the patches and Nhoj's cows):

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

int main() {
cin.tie(nullptr)->sync_with_stdio(false);
int K, M, N;
cin >> K >> M >> N;
vector<pair<int, int>> patches(K + M); // patches and Nhoj's cows
for (int i = 0; i < K; ++i)
cin >> patches[i].first >> patches[i].second;
for (int i = K; i < K + M; ++i) {
cin >> patches[i].first;
patches[i].second = -1;
}
sort(begin(patches), end(patches));
vector<uint64_t> increases;
int last_i = -1;
uint64_t sum_range = 0;
for (int i = 0; i < (int)patches.size(); ++i) {
if (patches[i].second == -1) {
if (last_i == -1) { // try placing to left of Nhoj's leftmost cow
increases.push_back(sum_range);
} else {
uint64_t cur_ans_1 = 0;	 // current sum of window
uint64_t best_ans_1 = 0; // best sum over all windows
for (int j = last_i + 1, r = last_i; j < i; ++j) {
// assume j is the leftmost patch in the window
while (r + 1 < i &&
(patches[r + 1].first - patches[j].first) * 2 <
patches[i].first - patches[last_i].first) {
cur_ans_1 += patches[++r].second;
}
// window sum is now sum(tastinesses(j..r))
best_ans_1 = max(best_ans_1, cur_ans_1);
cur_ans_1 -= patches[j].second;
}
assert(2 * best_ans_1 >= sum_range);
increases.push_back(best_ans_1);
increases.push_back(sum_range - best_ans_1);
}
last_i = i;
sum_range = 0;
} else {
sum_range += patches[i].second;
}
}
// try placing to right of Nhoj's rightmost cow
increases.push_back(sum_range);
// sort in decreasing order
sort(rbegin(increases), rend(increases));
increases.resize(N);
uint64_t ans = 0;
for (auto val : increases)
ans += val;
cout << ans << "\n";
}


Danny Mittal's code:

import java.io.BufferedReader;
import java.io.IOException;
import java.util.Arrays;
import java.util.Map;
import java.util.StringTokenizer;
import java.util.TreeMap;

public class ClosestCowWins {

public static void main(String[] args) throws IOException {
int k = Integer.parseInt(tokenizer.nextToken());
int m = Integer.parseInt(tokenizer.nextToken());
int n = Integer.parseInt(tokenizer.nextToken());
Patch[] patches = new Patch[k];
for (int j = 0; j < k; j++) {
int location = Integer.parseInt(tokenizer.nextToken());
long tastiness = Long.parseLong(tokenizer.nextToken());
patches[j] = new Patch(location, tastiness);
}
Arrays.sort(patches);
long[] tastinessSums = new long[k + 1];
for (int j = 0; j < k; j++) {
tastinessSums[j + 1] = tastinessSums[j] + patches[j].tastiness;
}
Integer[] nhojCows = new Integer[m];
for (int j = 0; j < m; j++) {
}
Arrays.sort(nhojCows);
TreeMap<Integer, Integer> nhojCowsTreeMap = new TreeMap<>();
for (int j = 0; j < m; j++) {
nhojCowsTreeMap.put(nhojCows[j], j);
}
TreeMap<Integer, Integer> patchTreeMap = new TreeMap<>();
for (int j = 0; j < k; j++) {
patchTreeMap.put(patches[j].location, j);
}
long[][] valueAdded = new long[3][m + 1];
for (int j = 0; j < k; j++) {
Map.Entry<Integer, Integer> nextNhojCowEntry = nhojCowsTreeMap.ceilingEntry(patches[j].location);
int nextNhojCow = nextNhojCowEntry == null ? m : nextNhojCowEntry.getValue();
if (nextNhojCow == 0 || nextNhojCow == m) {
} else {
int johnCowLocation = Math.min(nhojCows[nextNhojCow], (2 * patches[j].location) - nhojCows[nextNhojCow - 1]);
int extent = (johnCowLocation + nhojCows[nextNhojCow] + 1) / 2;
int farthestPatch = patchTreeMap.lowerEntry(extent).getValue();
}
}
Long[] valueAddedOverall = new Long[2 * (m + 1)];
for (int j = 0; j <= m; j++) {
}
for (int j = Math.max(0, valueAddedOverall.length - n); j < valueAddedOverall.length; j++) {
}
}

public static class Patch implements Comparable<Patch> {
final int location;
final long tastiness;

public Patch(int location, long tastiness) {
this.location = location;
this.tastiness = tastiness;
}

@Override
public int compareTo(Patch other) {
return location - other.location;
}
}
}


Bonus: It would still be possible to solve this question even if the condition $2\cdot (\text{one-cow answer})\ge (\text{two-cow answer})$ did not hold (see this problem).
The passage describes a problem involving Farmer Nhoj's cows and John's strategy to maximize the total tastiness of patches on a number line. The solution involves partitioning the number line into intervals, calculating the maximum tastiness John can claim with one or two cows in each interval, and then combining these results to find the optimal solution.
no_question

the passage does not contain a direct mathematical question or provide an answer to one. instead, it provides a detailed explanation of a problem and its solution, including code implementations.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
(Analysis by Benjamin Qi)

Farmer Nhoj's cows partition the number line into $M+1$ intervals. The general idea is to find the answers for each of these intervals independently, and then combine them.

Firstly, note that John can claim the total tastiness of the leftmost interval with a single cow. Similar reasoning holds for the rightmost interval. Every other of the $M-1$ intervals has one of Nhoj's cows at both of its endpoints, so one of John's cows might not be enough to claim all of the tastinesses within it. But placing two cows will definitely be enough (one just to the right of the left endpoint of the interval, and another just to the left of the right endpoint of the interval).

Assume all the patches and cows are sorted in increasing order of position, and consider the interval $(f_{x-1},f_x)$. Probably the trickiest part of the solution involves computing the maximum tastiness John can claim if he places only one cow in this interval. If John places a cow at $q\in (f_{x-1},f_x)$, he claims the tastinesses of all patches with positions in the range $\left(\frac{f_{x-1}+q}{2},\frac{f_{x}+q}{2}\right)$. Note that the length of this range ("window") is equal to $\frac{f_x-f_{x-1}}{2}$ regardless of $q$. We can slide this constant-length window from left to right and keep track of how the sum of tastinesses changes as patches enter and exit this window. The one-cow answer for this interval will equal the maximum sliding window sum that we ever encounter.

Now that we've computed the one-cow and two-cow answers for every interval, it remains to combine these to get the answer to the original problem. If John was restricted to placing at most one cow in every interval, then we could create a list of all the one-cow answers, sort it in decreasing order, and sum its first $N$ elements. It turns out that a slight modification of this strategy works when we allow John to place up to two cows in every interval: add the quantity $(\text{two-cow answer} - \text{one-cow answer})$ for each interval to the list before sorting it (if there are ties, put quantities of the form $\text{one-cow answer}$ before quantities of the form $(\text{two-cow answer} - \text{one-cow answer})$).

The reason these works is because $2\cdot (\text{one-cow answer})\ge (\text{two-cow answer})$ for any interval; in other words, adding a second cow to an interval can't increase the total tastiness more than adding the first cow did. This implies that if a prefix of the list contains the quantity $(\text{two-cow answer} - \text{one-cow answer})$ for an interval, it will also contain the quantity $\text{one-cow answer}$ for that same interval.

My code (with a single vector for both the patches and Nhoj's cows):

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

int main() {
cin.tie(nullptr)->sync_with_stdio(false);
int K, M, N;
cin >> K >> M >> N;
vector<pair<int, int>> patches(K + M); // patches and Nhoj's cows
for (int i = 0; i < K; ++i)
cin >> patches[i].first >> patches[i].second;
for (int i = K; i < K + M; ++i) {
cin >> patches[i].first;
patches[i].second = -1;
}
sort(begin(patches), end(patches));
vector<uint64_t> increases;
int last_i = -1;
uint64_t sum_range = 0;
for (int i = 0; i < (int)patches.size(); ++i) {
if (patches[i].second == -1) {
if (last_i == -1) { // try placing to left of Nhoj's leftmost cow
increases.push_back(sum_range);
} else {
uint64_t cur_ans_1 = 0;	 // current sum of window
uint64_t best_ans_1 = 0; // best sum over all windows
for (int j = last_i + 1, r = last_i; j < i; ++j) {
// assume j is the leftmost patch in the window
while (r + 1 < i &&
(patches[r + 1].first - patches[j].first) * 2 <
patches[i].first - patches[last_i].first) {
cur_ans_1 += patches[++r].second;
}
// window sum is now sum(tastinesses(j..r))
best_ans_1 = max(best_ans_1, cur_ans_1);
cur_ans_1 -= patches[j].second;
}
assert(2 * best_ans_1 >= sum_range);
increases.push_back(best_ans_1);
increases.push_back(sum_range - best_ans_1);
}
last_i = i;
sum_range = 0;
} else {
sum_range += patches[i].second;
}
}
// try placing to right of Nhoj's rightmost cow
increases.push_back(sum_range);
// sort in decreasing order
sort(rbegin(increases), rend(increases));
increases.resize(N);
uint64_t ans = 0;
for (auto val : increases)
ans += val;
cout << ans << "\n";
}


Danny Mittal's code:

import java.io.BufferedReader;
import java.io.IOException;
import java.util.Arrays;
import java.util.Map;
import java.util.StringTokenizer;
import java.util.TreeMap;

public class ClosestCowWins {

public static void main(String[] args) throws IOException {
int k = Integer.parseInt(tokenizer.nextToken());
int m = Integer.parseInt(tokenizer.nextToken());
int n = Integer.parseInt(tokenizer.nextToken());
Patch[] patches = new Patch[k];
for (int j = 0; j < k; j++) {
int location = Integer.parseInt(tokenizer.nextToken());
long tastiness = Long.parseLong(tokenizer.nextToken());
patches[j] = new Patch(location, tastiness);
}
Arrays.sort(patches);
long[] tastinessSums = new long[k + 1];
for (int j = 0; j < k; j++) {
tastinessSums[j + 1] = tastinessSums[j] + patches[j].tastiness;
}
Integer[] nhojCows = new Integer[m];
for (int j = 0; j < m; j++) {
}
Arrays.sort(nhojCows);
TreeMap<Integer, Integer> nhojCowsTreeMap = new TreeMap<>();
for (int j = 0; j < m; j++) {
nhojCowsTreeMap.put(nhojCows[j], j);
}
TreeMap<Integer, Integer> patchTreeMap = new TreeMap<>();
for (int j = 0; j < k; j++) {
patchTreeMap.put(patches[j].location, j);
}
long[][] valueAdded = new long[3][m + 1];
for (int j = 0; j < k; j++) {
Map.Entry<Integer, Integer> nextNhojCowEntry = nhojCowsTreeMap.ceilingEntry(patches[j].location);
int nextNhojCow = nextNhojCowEntry == null ? m : nextNhojCowEntry.getValue();
if (nextNhojCow == 0 || nextNhojCow == m) {
} else {
int johnCowLocation = Math.min(nhojCows[nextNhojCow], (2 * patches[j].location) - nhojCows[nextNhojCow - 1]);
int extent = (johnCowLocation + nhojCows[nextNhojCow] + 1) / 2;
int farthestPatch = patchTreeMap.lowerEntry(extent).getValue();
}
}
Long[] valueAddedOverall = new Long[2 * (m + 1)];
for (int j = 0; j <= m; j++) {
}
for (int j = Math.max(0, valueAddedOverall.length - n); j < valueAddedOverall.length; j++) {
}
}

public static class Patch implements Comparable<Patch> {
final int location;
final long tastiness;

public Patch(int location, long tastiness) {
this.location = location;
this.tastiness = tastiness;
}

@Override
public int compareTo(Patch other) {
return location - other.location;
}
}
}


Bonus: It would still be possible to solve this question even if the condition $2\cdot (\text{one-cow answer})\ge (\text{two-cow answer})$ did not hold (see this problem).
user
The passage describes a problem involving Farmer Nhoj's cows and John's strategy to maximize the total tastiness of patches on a number line. The solution involves partitioning the number line into intervals, calculating the maximum tastiness John can claim with one or two cows in each interval, and then combining these results to find the optimal solution.

TYPE: NO_QUESTION

The passage does not contain a direct mathematical question or provide an answer to one. Instead, it provides a detailed explanation of a problem and its solution, including code implementations.