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ACCENTURE EXAM SOLUTIONS

ACCENTURE EXAM SOLUTIONS

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πŸ”₯Guys plz Stop fearing for daily exams πŸ“ πŸ‘¨β€πŸ’» @srksvk is here to help you all at lowest cost possible.πŸ’ͺ πŸŒ€ ” Our Only Aim Is To Let Get Placed To You In A Reputed Company πŸ”₯Effort from our side = πŸ’― πŸ“±Main Channel: @coding_are πŸ“±Tel I'd : @srksvk

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πŸ“ˆ Analytical overview of Telegram channel ACCENTURE EXAM SOLUTIONS

Channel ACCENTURE EXAM SOLUTIONS (@coding_are) in the English language segment is an active participant. Currently, the community unites 14 129 subscribers, ranking 14 097 in the Education category and 28 067 in the India region.

πŸ“Š Audience metrics and dynamics

Since its creation on Π½Π΅Π²Ρ–Π΄ΠΎΠΌΠΎ, the project has demonstrated rapid growth, gathering an audience of 14 129 subscribers.

According to the latest data from 26 September, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -112 over the last 30 days and by -6 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 3.64%. Within the first 24 hours after publication, content typically collects 1.54% reactions from the total number of subscribers.
  • Post reach: On average, each post receives 514 views. Within the first day, a publication typically gains 218 views.
  • Reactions and interaction: The audience actively supports content: the average number of reactions per post is 2.
  • Thematic interests: Content is focused on key topics such as placement, gaurntee, suree, capgemini, infosy.

πŸ“ Description and content policy

The author describes the resource as a platform for expressing subjective opinions:
β€œπŸ”₯Guys plz Stop fearing for daily exams πŸ“ πŸ‘¨β€πŸ’» @srksvk is here to help you all at lowest cost possible.πŸ’ͺ πŸŒ€ ” Our Only Aim Is To Let Get Placed To You In A Reputed Company πŸ”₯Effort from our side = πŸ’― πŸ“±Main Channel: @coding_are πŸ“±Tel I'd : @srks...”

Thanks to the high frequency of updates (latest data received on 27 September, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Education category.

14 128
Subscribers
-624 hours
-127 days
-11230 days
Posts Archive
#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 ☺️

After 16k i will upload one more code answer with all tets caes passed πŸ”₯πŸ₯³πŸ₯³πŸ₯³ Show share fast guys

Fast guy's next answer are ready

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All are correct code with verified ☺️ So do correctly everyone

using System; using System.Collections.Generic; class Program { static void Main() { int N = int.Parse(Console.ReadLine()); var graph = new Dictionary>(); var indegrees = new Dictionary(); for (int i = 0; i < N; i++) { var edge = Console.ReadLine().Split(); string from = edge[0]; string to = edge[1]; if (!graph.ContainsKey(from)) graph[from] = new List(); graph[from].Add(to); if (!indegrees.ContainsKey(from)) indegrees[from] = 0; if (!indegrees.ContainsKey(to)) indegrees[to] = 0; indegrees[to]++; } var words = Console.ReadLine().Split(); var levels = new Dictionary(); var queue = new Queue(); foreach (var node in indegrees.Keys) { if (indegrees[node] == 0) { levels[node] = 1; // Root level is 1 queue.Enqueue(node); } } while (queue.Count > 0) { var current = queue.Dequeue(); int currentLevel = levels[current]; if (graph.ContainsKey(current)) { foreach (var neighbor in graph[current]) { if (!levels.ContainsKey(neighbor)) { levels[neighbor] = currentLevel + 1; queue.Enqueue(neighbor); } indegrees[neighbor]--; if (indegrees[neighbor] == 0 && !levels.ContainsKey(neighbor)) { queue.Enqueue(neighbor); } } } } int totalValue = 0; foreach (var word in words) { if (levels.ContainsKey(word)) { totalValue += levels[word]; } else { totalValue += -1; } } Console.Write(totalValue); } } String Puzzle C+

photo content
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photo content

int main() { string s; cin >> s; int n = s.length(), res = 0; vector<int> v(n); for (int i = 0; i < n; ++i) cin >> v[i]; int lw = s[0] - '0', lwv = v[0]; for (int i = 1; i < n; ++i) { if (s[i] - '0' == lw) { res += min(lwv, v[i]); lwv = max(lwv, v[i]); } else { lw = s[i] - '0'; lwv = v[i]; } } cout << res; return 0; } Form alternating string Change accordingly before submitting to avoid plagiarism @codeing_are

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Tcs code vitaa answer?πŸ‘‡πŸ‘‡πŸ‘‡πŸ‘‡πŸ‘‡

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