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give ans code in cpp

Criado em: 12 de julho de 2025

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Pergunta

a83c6ce4-8338-45ed-9684-ded97c5bff83.jpga83c6ce4-8338-45ed-9684-ded97c5bff83.jpg
4afb3e1d-01ac-4585-b3c5-0d4c0e55dec9.jpg4afb3e1d-01ac-4585-b3c5-0d4c0e55dec9.jpg
5350a811-3c76-457f-b5f1-dce102ef3b0f.jpg5350a811-3c76-457f-b5f1-dce102ef3b0f.jpg
29e3df6f-326a-4ef0-9000-61a875303c41.jpg29e3df6f-326a-4ef0-9000-61a875303c41.jpg
26184d52-fd1c-4084-b720-9f2b09868234.jpg26184d52-fd1c-4084-b720-9f2b09868234.jpg
0e070c34-1a0c-4c17-ac4a-149f92c38ff7.jpg0e070c34-1a0c-4c17-ac4a-149f92c38ff7.jpg

give ans code in cpp

Python

python
from PIL import Image
import matplotlib.pyplot as plt

img = Image.open('/mnt/data/0e070c34-1a0c-4c17-ac4a-149f92c38ff7.jpg')
plt.imshow(img)
plt.axis('off')

(-0.5, 719.5, 1279.5, -0.5)

Python

python
import numpy as np
# crop approximate area around center where text is
h,w = img.size[1], img.size[0]
# Actually img.size returns width,height
w, h = img.size
# crop top area where text lines
crop = img.crop((0, 120, w, 600))
plt.imshow(crop)
plt.axis('off')

(-0.5, 719.5, 479.5, -0.5)

Python

python
crop2 = img.crop((0, 500, w, 800))
plt.imshow(crop2)
plt.axis('off')

(-0.5, 719.5, 299.5, -0.5)

Python

python
# manually inspect crop2 section coordinates to see text numbers left half
crop2_left = img.crop((0, 520, 350, 850))
plt.imshow(crop2_left)
plt.axis('off')

(-0.5, 349.5, 329.5, -0.5)

Python

python
img2 = Image.open('/mnt/data/29e3df6f-326a-4ef0-9000-61a875303c41.jpg')
plt.imshow(img2)
plt.axis('off')

(-0.5, 719.5, 1279.5, -0.5)

Python

python
plt.imshow(img2.crop((0, 100, 550, 450)))
plt.axis('off')

(-0.5, 549.5, 349.5, -0.5)

Python

python
img3 = Image.open('/mnt/data/4afb3e1d-01ac-4585-b3c5-0d4c0e55dec9.jpg')
plt.imshow(img3)
plt.axis('off')

(-0.5, 719.5, 1279.5, -0.5)

Python

python
plt.imshow(img3.crop((0, 560, 550, 840)))
plt.axis('off')

(-0.5, 549.5, 279.5, -0.5)

Pensamento

Clarifying implementation

The algorithm is clear, so now focusing on implementation details. To store coordinates efficiently, I could use an unordered set of pairs, either as a long long encoded pair or as strings, but I think the pair of integers might be simpler. The coordinate range isn't specified, but assuming they are small, using a custom hash for pairs of integers seems appropriate. This will ensure quick lookups without excessive performance issues. Let's go with the pair-based unordered set for now.

Thought for 1m 41s

Resposta

C++ solution (axis-aligned squares, O(N²)):

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

struct PairHash {
    size_t operator()(const pair<int,int>& p) const noexcept {
        return (static_cast<uint64_t>(p.first) << 32) ^ static_cast<uint32_t>(p.second);
    }
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int N;
    if(!(cin >> N) || N < 4) return 0;          // no points ⇒ no output

    vector<pair<int,int>> pts(N);
    unordered_set<pair<int,int>, PairHash> S;   // constant-time lookup

    for (int i = 0; i < N; ++i) {
        cin >> pts[i].first >> pts[i].second;
        S.insert(pts[i]);
    }

    int best = INT_MAX;                         // smallest side found so far

    /* Enumerate every horizontal segment (two points with the same y).
       Treat it as the *bottom* edge, then also as the *top* edge (dir = ±1). */
    for (int i = 0; i < N; ++i) {
        for (int j = i + 1; j < N; ++j) {
            if (pts[i].second != pts[j].second) continue;   // not horizontal

            int side = abs(pts[i].first - pts[j].first);    // length of the edge
            if (side == 0) continue;                        // same point

            for (int dir : {+1, -1}) {                     // square above & below
                int newY = pts[i].second + dir * side;

                if (S.count({pts[i].first, newY}) &&
                    S.count({pts[j].first, newY})) {
                    best = min(best, side);
                }
            }
        }
    }

    /* If at least one square was found, best < INT_MAX */
    if (best != INT_MAX)
        cout << best << '\n';
    else
        cout << -1 << '\n';    // or handle as per problem-setter’s spec
    return 0;
}

How it works

  1. Store all points in an unordered_set for O(1) existence checks.
  2. Loop over every unordered pair of points that share the same y (i.e. they form a horizontal segment).
  3. The horizontal distance side = |x₂ – x₁| is a candidate side length.
    For an axis-aligned square that uses this segment as its bottom (or top) edge, the other two corners must be at
    (x₁, y ± side) and (x₂, y ± side).
  4. Check those two points in O(1) with the set; if present, update the current minimum side.
  5. After all pairs are tested, output the smallest side found (or -1 if no square exists).

With N ≤ 100 the double loop costs at most 4 995 checks, easily within limits.

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