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ACCENTURE EXAM SOLUTIONS

ACCENTURE EXAM SOLUTIONS

前往频道在 Telegram

🔥Guys plz Stop fearing for daily exams 📝 👨‍💻 @srksvk is here to help you all at lowest cost possible.💪 🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company 🔥Effort from our side = 💯 📱Main Channel: @coding_are 📱Tel I'd : @srksvk

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📈 Telegram 频道 ACCENTURE EXAM SOLUTIONS 的分析概览

频道 ACCENTURE EXAM SOLUTIONS (@coding_are) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 14 129 名订阅者,在 教育 类别中位列第 14 097,并在 印度 地区排名第 28 073 位。

📊 受众指标与增长动态

自 невідомо 创建以来,项目保持高速增长,吸引了 14 129 名订阅者。

根据 27 九月, 2026 的最新数据,频道保持稳定运转。过去 30 天订阅人数变化为 -105,过去 24 小时变化为 2,整体触达仍然可观。

  • 认证状态: 未认证
  • 互动率 (ER): 平均受众互动率为 3.64%。内容发布后 24 小时内通常能获得 1.54% 的反应,占订阅者总量。
  • 帖子覆盖: 每篇帖子平均可获得 515 次浏览,首日通常累积 217 次浏览。
  • 互动与反馈: 受众积极参与,单帖平均反应数为 2。
  • 主题关注点: 内容集中在 placement, gaurntee, suree, capgemini, infosy 等核心主题上。

📝 描述与内容策略

作者将该频道定位为表达主观观点的平台:
“🔥Guys plz Stop fearing for daily exams 📝 👨‍💻 @srksvk is here to help you all at lowest cost possible.💪 🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company 🔥Effort from our side = 💯 📱Main Channel: @coding_are 📱Tel I'd : @srks...”

凭借高频更新(最新数据采集于 28 九月, 2026),频道始终保持新鲜度与高覆盖。分析显示受众积极互动,使其成为 教育 类别中的关键影响点。

14 128
订阅者
+224 小时
-77 天
-10530 天
帖子存档
double area = calculateArea(lines); System.out.printf("%.2f\n", area); boolean canFormSameFigure = canRecreateShape(lines, area); System.out.println(canFormSameFigure ? "Yes" : "No"); } else { System.out.print("No"); } sc.close(); } }

import java.util.*; public class AaravAndArjun { static class Point { int x, y; Point(int x, int y) { this.x = x; this.y = y; } @Override public boolean equals(Object obj) { if (this == obj) return true; if (!(obj instanceof Point)) return false; Point p = (Point) obj; return x == p.x && y == p.y; } @Override public int hashCode() { return Objects.hash(x, y); } } static class Line { Point start, end; Line(int x1, int y1, int x2, int y2) { this.start = new Point(x1, y1); this.end = new Point(x2, y2); } Set<Point> getEndpoints() { Set<Point> endpoints = new HashSet<>(); endpoints.add(start); endpoints.add(end); return endpoints; } } private static boolean isClosedFigure(List<Line> lines) { Map<Point, Integer> endpointCount = new HashMap<>(); // Count the occurrences of each endpoint for (Line line : lines) { Set<Point> endpoints = line.getEndpoints(); for (Point point : endpoints) { endpointCount.put(point, endpointCount.getOrDefault(point, 0) + 1); } } // There must be an even count of endpoints for a closed figure for (int count : endpointCount.values()) { if (count % 2 != 0) { return false; } } return endpointCount.size() >= 3; // Must be at least a triangle } private static double calculateArea(List<Line> lines) { // Assuming the lines form a simple polygon, we can use the shoelace formula double area = 0.0; List<Point> vertices = new ArrayList<>(); // Collect vertices of the polygon from lines for (Line line : lines) { vertices.add(line.start); vertices.add(line.end); } // Remove duplicate vertices Set<Point> uniqueVertices = new HashSet<>(vertices); List<Point> vertexList = new ArrayList<>(uniqueVertices); // Sort vertices in a counter-clockwise manner around the centroid // (not implemented for simplicity) int n = vertexList.size(); for (int i = 0; i < n; i++) { Point p1 = vertexList.get(i); Point p2 = vertexList.get((i + 1) % n); area += p1.x * p2.y - p2.x * p1.y; } return Math.abs(area) / 2.0; } private static boolean canRecreateShape(List<Line> lines, double area) { // Calculate total length of leftover sticks and compare with perimeter of the shape double leftoverLength = 0.0; for (Line line : lines) { double length = Math.sqrt(Math.pow(line.end.x - line.start.x, 2) + Math.pow(line.end.y - line.start.y, 2)); leftoverLength += length; } // For this problem, we assume we can recreate the shape if we have enough leftover length return leftoverLength >= area; // This is a simplification } public static void main(String[] args) { Scanner sc = new Scanner(System.in); int N = sc.nextInt(); // Number of sticks List<Line> lines = new ArrayList<>(); for (int i = 0; i < N; i++) { int x1 = sc.nextInt(); int y1 = sc.nextInt(); int x2 = sc.nextInt(); int y2 = sc.nextInt(); lines.add(new Line(x1, y1, x2, y2)); } if (isClosedFigure(lines)) { System.out.println("Yes");

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import java.util.HashMap; import java.util.Map; import java.util.Scanner; public class MinimumMoves { public static int minimumMoves(String[] instruct) { String[] directions = {"up", "down", "left", "right"}; int s = instruct.length; int[][][] dp = new int[s + 1][4][4]; for (int i = 0; i <= s; i++) { for (int j = 0; j < 4; j++) { for (int k = 0; k < 4; k++) { dp[i][j][k] = Integer.MAX_VALUE; } } } for (int i = 0; i < 4; i++) { for (int j = 0; j < 4; j++) { dp[0][i][j] = 0; } } for (int k = 1; k <= s; k++) { int instrIdx = -1; for (int i = 0; i < 4; i++) { if (directions[i].equals(instruct[k - 1])) { instrIdx = i; break; } } for (int i = 0; i < 4; i++) { for (int j = 0; j < 4; j++) { if (dp[k - 1][i][j] != Integer.MAX_VALUE) { if (instrIdx == i || instrIdx == j) { dp[k][i][j] = Math.min(dp[k][i][j], dp[k - 1][i][j]); } else { dp[k][i][j] = Math.min(dp[k][i][j], dp[k - 1][instrIdx][j] + 1); dp[k][i][j] = Math.min(dp[k][i][j], dp[k - 1][i][instrIdx] + 1); } } } } } int minimumMoves = Integer.MAX_VALUE; for (int i = 0; i < 4; i++) { for (int j = 0; j < 4; j++) { minimumMoves = Math.min(minimumMoves, dp[s][i][j]); } } return minimumMoves; } public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int m = Integer.parseInt(scanner.nextLine().trim()); String[] instruct = new String[m]; for (int i = 0; i < m; i++) { instruct[i] = scanner.nextLine().trim(); } int result = minimumMoves(instruct); System.out.print(result); scanner.close(); } } Dance rev Java 8

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#include <bits/stdc++.h> using namespace std; #define loop(i, a, n) for (lli i = (a); i < (n); ++i) #define loopD(i, a, n) for (lli i = (a); i >= (n); --i) #define all(c) (c).begin(), (c).end() #define rall(c) (c).rbegin(), (c).rend() #define sz(a) ((lli)a.size()) #define YES cout << "YES" << endl; #define NO cout << "NO" << endl; #define endl '\n' #define fastio std::ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL); #define pb push_back #define pp pop_back() #define fi first #define si second #define v(a) vector<int>(a) #define vv(a) vector<vector<int>>(a) #define present(c, x) ((c).find(x) != (c).end()) #define set_bits __builtin_popcountll #define MOD 1000000007 #define int long long typedef long long lli; typedef vector<int> vi; typedef vector<vi> vvi; typedef pair<lli, lli> pll; typedef pair<int, int> pii; typedef unordered_map<int, int> umpi; typedef map<int, int> mpi; typedef vector<pii> vp; typedef vector<lli> vll; typedef vector<vll> vvll; struct Line { int x1, y1, x2, y2; bool vertical() const { return x1 == x2; } bool horizontal() const { return y1 == y2; } bool diagonal() const { return abs(x2 - x1) == abs(y2 - y1); } }; int N, K; vector<Line> lines; map<pair<int, int>, vector<int>> ptsMap; void add(const Line& line, int idx) { int x1 = line.x1, y1 = line.y1; int x2 = line.x2, y2 = line.y2; if (line.vertical()) { int yStart = min(y1, y2); int yEnd = max(y1, y2); for(int y = yStart; y <= yEnd; y++) { ptsMap[{x1, y}].push_back(idx); } } else if (line.horizontal()) { int xStart = min(x1, x2); int xEnd = max(x1, x2); for(int x = xStart; x <= xEnd; x++) { ptsMap[{x, y1}].push_back(idx); } } else if (line.diagonal()) { int steps = abs(x2 - x1); int dx = (x2 - x1) / steps; int dy = (y2 - y1) / steps; for(int i = 0; i <= steps; i++) { int x = x1 + i * dx; int y = y1 + i * dy; ptsMap[{x, y}].push_back(idx); } } } int ff(int x1, int y1, int x2, int y2) { if(x1 == x2) return abs(y1 - y2); if(y1 == y2) return abs(x1 - x2); if(abs(x1 - x2) == abs(y1 - y2)) return abs(x1 - x2); return 0; } int solve(const pair<int, int>& pt, const vector<int>& lst) { vector<int> d; for(auto lIdx : lst) { const Line& ln = lines[lIdx]; bool oneSided = (pt.first == ln.x1 && pt.second == ln.y1) || (pt.first == ln.x2 && pt.second == ln.y2); if(oneSided) { int ex = (pt.first == ln.x1 && pt.second == ln.y1) ? ln.x2 : ln.x1; int ey = (pt.first == ln.x1 && pt.second == ln.y1) ? ln.y2 : ln.y1; d.push_back(ff(pt.first, pt.second, ex, ey)); } else { d.push_back(ff(pt.first, pt.second, ln.x1, ln.y1)); d.push_back(ff(pt.first, pt.second, ln.x2, ln.y2)); } } return d.empty() ? 0 : *min_element(d.begin(), d.end()); } void solve() { cin >> N; lines.resize(N); for(int i = 0; i < N; i++) { cin >> lines[i].x1 >> lines[i].y1 >> lines[i].x2 >> lines[i].y2; add(lines[i], i); } cin >> K; int total = 0; for(auto &[pt, lst] : ptsMap) { if(sz(lst) == K) { total += solve(pt, lst); } } cout << total; } int32_t main() { solve(); return 0; } // magic stars intensity ac Full accepted ☺️

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Changes variable

Full accepted ☺️

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