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Nobita's neighborhood is under attack by Gian's po...

Créé le : 13 juillet 2025

Répondu en utilisant O4 Mini High par Chat01

Question

Nobita's neighborhood is under attack by Gian's powerful robot army.

The only defense is a long protective barrier, which is divided into N sections Each section already has some security cameras installed, which can detect robots at nearby sections.

A security camera at section i can scan all sections j within a distance of at most R meaning it can help in monitoring sections where li - jl ≤ R

The Shield Level of a section is the total number of cameras (including those from neighboring sections) monitoring it.

To stop the robot invasion, Doraemon has K additional cameras that he can place anywhere along the barrier.

The goal is to maximize the weakest section's defense level, ensuring that no section is left vulnerable.

Find the maximum possible defense level for the weakest section.

Input Format

The first line contains an integer, N, denoting the number of sections in the barrier

The next line contains an integer, R, denoting the maximum range a camera can scan.
The next line contains an integer, K, denoting the number of additional cameras available.

Each line i of the N subsequent lines (where 1 <= i <= N ) contains an integer describing Cameras[i].

Constraints

1 <= N <= 10 ^ 5

<= R <= N

0 <= K <= 10 ^ 12

0 <= Cameras[i] <= 10^9

Sample Test Cases

Case 1

Input
Input

5

03534 480

Output:

5

Explanation:

N-5 R-0 K-3

Cameras [534891

Explanation:

Each camera only monitors its own section (R0).

Current weakest section: The 2nd section (only 3 cameras).

We place 2 extra cameras there defense level becomes 5.

Distribute the remaining 1 cameras optimally to other weaker sections ile 3rd one

The minimum defense level across all sections is now 5.
code in java without comments and buffer reader

Réflexion

Breaking down the problem

The user is asking about an algorithmic programming problem. The goal is to determine how to distribute K cameras across N sections to maximize the minimum shield level. We can solve this using binary search on a target T, which represents the minimum shield level. Then, we use a sliding window sum to compute existing coverage, and a difference array to track the effect of adding cameras. It's a classic "maximize the minimum" type of problem!

Considering camera placement strategy

I need to figure out the best place to add cameras for maximum coverage. For each section, if its shield is below T, I’ll add cameras at the farthest right possible, but not beyond the last section. As I do this, I’ll track the cumulative effect of added cameras using a queue or difference array. I also need to work through the sample input, where it seems like some numbers might be incorrect. The goal is to use binary search to find the optimal shield level for each section.

Optimizing the camera placement algorithm

I’m working through the logic for placing the cameras to maximize coverage. The plan is to use binary search with a target mid value. For each section, I’ll add cameras to cover its deficit, ensuring that the effect starts at the right place by calculating the range it will impact. I’ll maintain two arrays: one for the additional coverage depending on positions and another for when coverage ends. All of this relies on precomputing the initial shield level using a sliding window sum. I'll implement the logic in Java while following the input/output requirements.

Thought for 13 seconds

Réponse

java
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int R = sc.nextInt();
        long K = sc.nextLong();
        long[] cams = new long[N];
        for (int i = 0; i < N; i++) {
            cams[i] = sc.nextLong();
        }
        long[] prefix = new long[N];
        prefix[0] = cams[0];
        for (int i = 1; i < N; i++) {
            prefix[i] = prefix[i - 1] + cams[i];
        }
        long[] init = new long[N];
        for (int i = 0; i < N; i++) {
            int L = Math.max(0, i - R);
            int Rg = Math.min(N - 1, i + R);
            init[i] = prefix[Rg] - (L > 0 ? prefix[L - 1] : 0);
        }
        long low = Long.MAX_VALUE;
        for (long v : init) if (v < low) low = v;
        long high = low + K;
        while (low < high) {
            long mid = (low + high + 1) >>> 1;
            if (canReach(init, N, R, K, mid)) {
                low = mid;
            } else {
                high = mid - 1;
            }
        }
        System.out.println(low);
    }

    static boolean canReach(long[] init, int N, int R, long K, long target) {
        long used = 0, extra = 0;
        long[] endAt = new long[N + 1];
        for (int i = 0; i < N; i++) {
            extra -= endAt[i];
            long cur = init[i] + extra;
            if (cur < target) {
                long need = target - cur;
                used += need;
                if (used > K) return false;
                extra += need;
                int p = Math.min(N - 1, i + R);
                int e = Math.min(N - 1, p + R);
                endAt[e + 1] += need;
            }
        }
        return true;
    }
}

Question

You are given in three integers N. L and R

It is given that a flip flop array is delined as an array of length N such that:

Each element is between Land R

There are no two adjacent values that are equal

No three adjacent values that are increasing or decreasing

Moreover, it is guaranteed that L < R.

Find the total number of possible flip top arrays of size N. Since the answer can be large return it modulo 10^9+7.

Input Format

The first fine contains an integer. N. denoting the length of the array

The next line contains an integet, L. denoting the lower bound.

The next line contains an integer, R. denoting the upper bound

Constraints
3 <= N <= 2000

1 <= Lc = 2000

1 <= R <= 2000

Sample Test Cases

Case 1

Input:

3 1 2

Output

2

Explanation:

N=3, L-1, R-2

The arrays are:

[1, 2, 1]

17, 1, 21

Because for all i, there is no i such that A(i) = A(1 + 1)

There is no i such that [1] subset A [1 + 1] subset A[1+] ]

There is no i such that A[1] > A[1 + 1] ,>A[1-2]
Case: 2

Input:

333

Output:

10

Explanation:

N-3, L-1, R-3

The arrays are:

  1. [1, 2, 1]

  2. [1, 3, 1]

  3. [1, 3, 2]

  4. [2, 1, 2]

  5. [2, 1, 3]

  6. (2, 3, 1]

  7. [2, 3, 2]

  8. [3, 1, 2]

  9. (3, 1, 3]

  10. [3, 2, 3]1

Case 3

Input:

334

Output:

2

code in java without comments and buffer reader such that passes all test cases

Réflexion

Optimizing counting problem

The task involves counting arrays where no two adjacent elements are the same and no three adjacent elements strictly increase or decrease. The approach involves breaking down the problem with dynamic programming (DP) to track differences between adjacent elements and trends (increasing, decreasing, or none). The key challenge stems from dealing with large inputs: N and M go up to 2000, meaning a direct DP approach might be too slow. Refining the algorithm to track differences and trends efficiently is crucial.

Refining the sequence condition

The sequence needs to avoid strictly increasing or decreasing runs of length 3, but allows monotonic runs of length 2. To track this, the state should depend on the previous two values and their differences. A dynamic programming approach could work, but considering the constraints (N and M up to 2000), the size may be too large.

Instead, I’m considering a combinatorial approach. Sequences where differences alternate sign can achieve the desired "zigzag" property without violating the given conditions.

Defining zigzag sequences

To count all valid sequences, we look for zigzag patterns where adjacent differences alternate in sign and are never zero. These sequences can repeat non-adjacent elements but must have distinct adjacent values. The idea is to start with a particular direction for the second element (either increasing or decreasing), and every subsequent step alternates in sign. To compute the count of such sequences of length N over a set of M values, we use dynamic programming, considering both possible alternating sign patterns (starting with either "up" or "down").

Thought for 25 seconds

Réponse

java
import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int L = sc.nextInt();
        int R = sc.nextInt();
        int M = R - L + 1;
        int mod = 1000000007;
        System.out.println((count(N, M, mod, 1) + count(N, M, mod, -1)) % mod);
    }
    static int count(int N, int M, int mod, int p2) {
        int[] dpPrev = new int[M + 1];
        int[] dpCurr = new int[M + 1];
        int[] prefix = new int[M + 1];
        for (int i = 1; i <= M; i++) dpPrev[i] = 1;
        for (int i = 2; i <= N; i++) {
            prefix[0] = 0;
            for (int j = 1; j <= M; j++) {
                int v = dpPrev[j] + prefix[j - 1];
                if (v >= mod) v -= mod;
                prefix[j] = v;
            }
            boolean less = (p2 == 1 ? i % 2 == 0 : i % 2 == 1);
            for (int x = 1; x <= M; x++) {
                if (less) {
                    dpCurr[x] = prefix[x - 1];
                } else {
                    int v = prefix[M] - prefix[x];
                    if (v < 0) v += mod;
                    dpCurr[x] = v;
                }
            }
            int[] t = dpPrev; dpPrev = dpCurr; dpCurr = t;
        }
        int res = 0;
        for (int i = 1; i <= M; i++) {
            res += dpPrev[i];
            if (res >= mod) res -= mod;
        }
        return res;
    }
}

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