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Question

WhatsApp Image 2025-12-13 at 17.50.12_e63057a3.jpgWhatsApp Image 2025-12-13 at 17.50.12_e63057a3.jpg
WhatsApp Image 2025-12-13 at 17.48.45_7174b232.jpgWhatsApp Image 2025-12-13 at 17.48.45_7174b232.jpg

code in cpp

Thought

Interpreting task about chocolate boxes gifts

The task is asking for the maximum number of chocolate boxes the teacher can use in gifts, where each gift consists of two boxes and the sum of chocolates in those boxes is divisible by the number of students. The goal is to maximize pairs of boxes that satisfy the condition (a_i + a_j) % x == 0.

Explaining algorithm for maximizing box pairs

The task is to compute the maximum number of chocolate boxes the teacher can use in gifts. The algorithm counts how many boxes fall under each remainder when divided by x. It pairs boxes with matching remainders and handles edge cases where remainders can pair within themselves. The final complexity of this process is O(n + x).

Thought for 35s

Answer

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

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

    int T;
    cin >> T;
    while (T--) {
        int n, x;
        cin >> n >> x;

        vector<long long> cnt(x, 0);
        for (int i = 0; i < n; i++) {
            long long a;
            cin >> a;
            cnt[a % x]++;
        }

        long long pairs = 0;

        // remainder 0 pairs with remainder 0
        pairs += cnt[0] / 2;

        // r pairs with (x-r)
        for (int r = 1; r <= (x - 1) / 2; r++) {
            pairs += min(cnt[r], cnt[x - r]);
        }

        // if x even, remainder x/2 pairs with itself
        if (x % 2 == 0) {
            pairs += cnt[x / 2] / 2;
        }

        // answer = maximum number of boxes used (2 boxes per gift)
        cout << pairs * 2 << "\n";
    }
    return 0;
}

Question

WhatsApp Image 2025-12-13 at 17.51.28_035cffb2.jpgWhatsApp Image 2025-12-13 at 17.51.28_035cffb2.jpg
WhatsApp Image 2025-12-13 at 17.51.41_f59b08eb.jpgWhatsApp Image 2025-12-13 at 17.51.41_f59b08eb.jpg

code in cpp

Python

python
from PIL import Image
img1 = Image.open('/mnt/data/WhatsApp Image 2025-12-13 at 17.51.28_035cffb2.jpg')
img2 = Image.open('/mnt/data/WhatsApp Image 2025-12-13 at 17.51.41_f59b08eb.jpg')
img1.size, img2.size


((1280, 576), (1280, 576))

Python

python
import matplotlib.pyplot as plt, numpy as np
def show_crop(img, box):
    crop=img.crop(box)
    plt.figure(figsize=(10,4))
    plt.imshow(crop)
    plt.axis('off')
    plt.show()

# crop sample input/output area from img1
show_crop(img1, (0,250,1280,520))

Thought

Validating palindrome generation

For "adskassda," I've computed the frequencies and matched the output length (7). The key is forming the left half from pairs and picking the smallest odd character ('a') for the center. The palindrome that results is "adsasda," which matches the sample. For "talent," I noticed a discrepancy in character counts, but the sample output ("tat") seems to imply letter rearranging, which I can’t assume is allowed from the problem description.

Validating "talent" palindrome

I realized that "talent" does indeed have two 't's. After counting, I found that both 't's form a valid pair for the sides, and the smallest odd character 'a' is chosen for the center. Hence, the resulting palindrome is "tat," which matches. For "decrypt," since no pairs exist, 'c' is selected as the center. Everything checks out, so my algorithm seems correct—time to implement it in C++!

Thought for 1m 22s

Answer

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

static string solve_one(const string &s) {
    int cnt[26] = {0};
    for (char c : s) cnt[c - 'a']++;

    string left;
    left.reserve(s.size() / 2);

    // Build left half in increasing order for lexicographically smallest palindrome
    for (int i = 0; i < 26; i++) {
        left.append(cnt[i] / 2, char('a' + i));
    }

    // Choose smallest possible middle character (if any odd count exists)
    char mid = 0;
    for (int i = 0; i < 26; i++) {
        if (cnt[i] % 2 == 1) {
            mid = char('a' + i);
            break;
        }
    }

    string right = left;
    reverse(right.begin(), right.end());

    if (mid) return left + string(1, mid) + right;
    return left + right;
}

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

    int T;
    cin >> T;
    while (T--) {
        string s;
        cin >> s;
        cout << solve_one(s) << "\n";
    }
    return 0;
}

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