pair<double, double> r(double px, double py, double x1, double y1, double x2, double y2) {
    double a = y2 - y1;
    double b = x1 - x2;
    double c = x2 * y1 - x1 * y2;
    double d = (a * px + b * py + c) / sqrt(a * a + b * b);
    double nx = px - 2 * d * (a / sqrt(a * a + b * b));
    double ny = py - 2 * d * (b / sqrt(a * a + b * b));
    return {nx, ny};
}

int main() {
    double ar;
    cin >> ar;

    double x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;

    double s = sqrt(ar);
    vector<pair<double, double>> cr = {
        {0, 0},
        {0, s},
        {s, s},
        {s, 0},
    };

    set<pair<double, double>> pts(cr.begin(), cr.end());

    for (const auto& c : cr) {
        auto [rx, ry] = r(c.first, c.second, x1, y1, x2, y2);
        pts.insert({rx, ry});
    }

    for (const auto& p : pts) {
        cout << fixed << setprecision(2) << p.first << " " << p.second << endl;
    }

    return 0;
}

C++
Folded area - Codevita
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struct P {
    double a, b;
    P(double a = 0, double b = 0) : a(a), b(b) {}
};

P r(const P &p, double t) {
    return P(p.a * cos(t) - p.b * sin(t),
             p.a * sin(t) + p.b * cos(t));
}

pair<double, double> g(const vector<P> &q, double t) {
    double x1 = 1e9, x2 = -1e9, y1 = 1e9, y2 = -1e9;
    for (const auto &p : q) {
        P s = r(p, t);
        x1 = min(x1, s.a);
        x2 = max(x2, s.a);
        y1 = min(y1, s.b);
        y2 = max(y2, s.b);
    }
    return {x2 - x1, y2 - y1};
}

int main() {
    int n;
    cin >> n;
    vector<P> q(n);

    for (int i = 0; i < n; ++i) {
        cin >> q[i].a >> q[i].b;
    }

    double m = 1e9, w = 0, h = 0;

    for (int i = 0; i < 360; ++i) {
        double t = i * M_PI / 180.0;
        auto [cw, ch] = g(q, t);
        double a = cw * ch;
        if (a < m) {
            m = a;
            w = cw;
            h = ch;
        }
    }

    if (w > h) {
        swap(w, h);
    }

    cout << fixed << setprecision(0) << round(w) << " "
         << fixed << setprecision(0) << round(h);

    return 0;
}

C++
Folder Area - Codevita
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SELECT
    RoomType,
    SUM(BedCount) AS 'Total Beds'
FROM
    Rooms
GROUP BY
    RoomType;

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TCS NQT 2025

🎖Batch -        2020 - 2025

Test Date : 12th May 2025
          
https://www.tcsion.com/hub/national-qualifier-test/

WhatsApp -   https://bit.ly/3tcjxV3

Navi Off Campus Drive


Job Profile : AI Solution Engineer 1


🎖Batch -        2022  2023  2024  2025

           https://careers.navi.com/navi/jobview/ai-solutions-engineer-1-bangalore-karnataka-india-2025040217480958?source=linkedin


WhatsApp -   https://bit.ly/3tcjxV3

TCS NQT 2025 | Full Mock Test PDF
🔥 30+ High-Quality Questions | ✅ With Answers
📊 Numerical | ✍️ Verbal | 🧠 Reasoning | 💻 Coding

PDF Based on Real Exam Pattern
Boost your prep with the most expected questions!

Shared by: STUDY MATERIAL - Placement Jobs & Materials
Join for More Materials & Updates:
➡️ https://t.me/studymaterial_IT

def update_list(val, items=[]):
    items.append(val)
    return items

print("Call 1:", update_list(10))  
print("Call 2:", update_list(20))  

def fresh_list(val, items=None):
    if items is None:
        items = []
    items.append(val)
    return items

print("Call 3:", fresh_list(30))  
print("Call 4:", fresh_list(40))

Options:
A.
Call 1: [10]
Call 2: [10, 20]
Call 3: [30]
Call 4: [30, 40]

B.
Call 1: [10]
Call 2: [20]
Call 3: [30]
Call 4: [40]

C.
Call 1: [10]
Call 2: [10, 20]
Call 3: [30]
Call 4: [30]

D.
Error due to list mutation

Answer: A

🔥 PLACEMENT PREP COMBO PACK 🔥
Boost your logic + aptitude + coding skills in 5 mins!

1. Aptitude – Profit & Loss

A shopkeeper buys an item for Rs. 240 and sells it at a profit of 25%.
Selling Price = ?

Solution:
SP = CP × (1 + Profit%)
= 240 × (1 + 25/100)
= 240 × 1.25 = Rs. 300


2. Python Concept – List vs Tuple

List: Mutable
Example: my_list = [1, 2, 3]
You can modify → my_list[0] = 10

Tuple: Immutable
Example: my_tuple = (1, 2, 3)
You cannot modify elements.


3. Mini Coding Task – Count Vowels in String

def count_vowels(s): return sum(1 for ch in s.lower() if ch in 'aeiou') # Example: count_vowels("placement") → 

Output: 3

🔥Problem Statement🔥


A robot is stuck in a maze represented by an NxN grid. The grid contains:

'S': Starting point

'E': Ending point

'.': Empty cell (can move)

'#': Wall (cannot move)

'P': Portal (teleports to the next portal clockwise)

Rules:

Robot can move up, down, left, right.

If robot steps on 'P', it is instantly teleported to the next portal (clockwise).

You need to find the minimum number of steps to reach 'E' from 'S'.

#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;

typedef pair<int, int> pii;

int N;
vector<vector<char>> grid;
vector<pii> portals;

bool isValid(int x, int y, vector<vector<bool>>& visited) {
    return x >= 0 && x < N && y >= 0 && y < N && !visited[x][y] && grid[x][y] != '#';
}

pii getNextPortal(pii current) {
    for (int i = 0; i < portals.size(); i++) {
        if (portals[i] == current) {
            return portals[(i + 1) % portals.size()];
        }
    }
    return current;
}

int shortestPathWithPortals() {
    pii start, end;

    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            if (grid[i][j] == 'S') start = {i, j};
            else if (grid[i][j] == 'E') end = {i, j};
            else if (grid[i][j] == 'P') portals.push_back({i, j});
        }
    }

    vector<vector<bool>> visited(N, vector<bool>(N, false));
    queue<pair<pii, int>> q;

    q.push({start, 0});
    visited[start.first][start.second] = true;

    int dx[] = {-1, 1, 0, 0};
    int dy[] = {0, 0, -1, 1};

    while (!q.empty()) {
        pii curr = q.front().first;
        int steps = q.front().second;
        q.pop();

        if (curr == end) return steps;

        for (int d = 0; d < 4; d++) {
            int nx = curr.first + dx[d];
            int ny = curr.second + dy[d];

            if (isValid(nx, ny, visited)) {
                if (grid[nx][ny] == 'P') {
                    pii tele = getNextPortal({nx, ny});
                    if (!visited[tele.first][tele.second]) {
                        visited[tele.first][tele.second] = true;
                        q.push({tele, steps + 1});
                    }
                } else {
                    visited[nx][ny] = true;
                    q.push({{nx, ny}, steps + 1});
                }
            }
        }
    }

    return -1;
}

int main() {
    cin >> N;
    grid = vector<vector<char>>(N, vector<char>(N));
    for (int i = 0; i < N; i++)
        for (int j = 0; j < N; j++)
            cin >> grid[i][j];

    cout << shortestPathWithPortals() << endl;
    return 0;
}

C++

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