uz
Feedback

Firibgarlarga uchmang! Telemetrio bunday kanallarni topadi va belgilaydi 👉 Belgini ko‘rmoqchi bo‘lsangiz, obuna bo‘ling

ACCENTURE EXAM SOLUTIONS

ACCENTURE EXAM SOLUTIONS

Kanalga Telegram’da o‘tish

🔥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

Ko'proq ko'rsatish

📈 Telegram kanali ACCENTURE EXAM SOLUTIONS analitikasi

ACCENTURE EXAM SOLUTIONS (@coding_are) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 14 039 obunachidan iborat bo'lib, Taʼlim toifasida 14 513-o'rinni va Hindiston mintaqasida 28 403-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 14 039 obunachiga ega bo‘ldi.

14 Oktabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -133 ga, so‘nggi 24 soatda esa -6 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 3.61% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.82% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 507 marta ko‘riladi; birinchi sutkada odatda 256 ta ko‘rish yig‘iladi.
  • Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 2 ta reaksiya keladi.
  • Tematik yo‘nalishlar: Kontent placement, gaurntee, suree, capgemini, infosy kabi asosiy mavzularga jamlangan.

📝 Tavsif va kontent siyosati

Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
“🔥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...”

Yuqori yangilanish chastotasi (oxirgi ma’lumot 15 Oktabr, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Taʼlim toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.

14 039
Obunachilar
-624 soatlar
-347 kun
-13330 kun
Postlar arxiv
#include <iostream> #include <vector> using namespace std; int main() { int n; cin >> n; vector<tuple<int, int, int>> stocks; for (int i = 0; i < n; ++i) { int a, b, c; cin >> a >> b >> c; stocks.push_back(make_tuple(a, b, c)); } int m; cin >> m; int real = 0; int unreal = 0; vector<vector<int>> prices; for (int i = 0; i < n; ++i) { vector<int> arr(m); for (int j = 0; j < m; ++j) { cin >> arr[j]; } prices.push_back(arr); } int day; cin >> day; for (int i = 0; i < n; ++i) { int a, b, c; tie(a, b, c) = stocks[i]; if (b > day) { continue; } else if (c > day || !c) { unreal += a * (prices[i][day - 1] - prices[i][b - 1]); } else { real += a * (prices[i][c - 1] - prices[i][b - 1]); } } cout << real << endl; cout << unreal << endl; return 0; } Roi code fully passsd ✅✅✅✅✅ Share @codeing_area

def toggle(num): result = "" for bit in num: if bit == "0": result += "1" else: result += "0" return result def get_max_sum(arr, k): max_val = max(arr) max_index = arr.index(max_val) left = max(0, max_index-k) right = min(len(arr)-1, max_index+k) selected = arr[left:max_index+1] + arr[max_index:right+1] arr = [x for x in arr if x not in selected] return sum(selected), arr # Input arr = list(map(int, input().split())) a1, b1 = input().split() a2, b2 = input().split() k = int(input()) sum1, sum2 = 0, 0 while arr: s1, arr = get_max_sum(arr, k) sum1 += s1 if not arr: break s2, arr = get_max_sum(arr, k) sum2 += s2 if sum1 > sum2: a1 = toggle(a1) b2 = toggle(b2) else: a2 = toggle(a2) b1 = toggle(b1) xor1 = int(a1, 2) ^ int(b1, 2) xor2 = int(a2, 2) ^ int(b2, 2) if xor1 > xor2: print("Rahul",end="") elif xor2 > xor1: print("Rupesh",end="") else: print("both",end="") Who is Lucky

def hamming_distance(str1, str2): return sum(c1 != c2 for c1, c2 in zip(str1, str2)) def minimize_cost_and_print_distance(binary_string, a, b): # Count occurrences of "01" and "10" count_01 = binary_string.count("01") count_10 = binary_string.count("10") # Calculate costs cost_01 = count_01 * a cost_10 = count_10 * b # Determine rearranged string for minimum cost if cost_01 < cost_10: rearranged_string = "01" * count_01 + "0" * (len(binary_string) - 2 * count_01) else: rearranged_string = "10" * count_10 + "1" * (len(binary_string) - 2 * count_10) # Calculate and print hamming distance distance = hamming_distance(binary_string, rearranged_string) print(distance) # Input reading and processing T = int(input("Enter the number of test cases: ")) for _ in range(T): binary_string = input("Enter the binary string: ") a, b = map(int, input("Enter A and B separated by space: ").split()) minimize_cost_and_print_distance(binary_string, a, b) Hamming distance Share @codeing_area

Tcs code vitaaa passed successfully ✅✅✅ Share everyone i will post soon
Tcs code vitaaa passed successfully ✅✅✅ Share everyone i will post soon

Share our channel One more code soon Need your support guys Sitting from last 3 hours to share Codes in free Share it @codeing_area

Hlo everyone share group only ✅✅✅ Share group --- @codeing_area

ROI 100% Working ✅ Python TCS Codevita @codeing_area n = int(input()) stocks = [] for _ in range(n): a, b, c = map(int, input().split()) stocks.append((a, b, c)) m = int(input()) real = 0 unreal = 0 placementlelo = [] for i in range(n): arr = [int(i) for i in input().split()] placementlelo.append(arr) day = int(input()) for i in range(n): a, b, c = stocks[i] if b > day: continue elif c > day or not c: unreal += a * (placementlelo[i][day - 1] - placementlelo[i][b - 1]) else: real += a * (placementlelo[i][c - 1] - placementlelo[i][b - 1]) print(real) print(unreal) ROI 100% Working ✅ Python TCS Codevita @codeing_area

def toggle(num): result = "" for bit in num: if bit == "0": result += "1" else: result += "0" return result def get_max_sum(arr, k): max_val = max(arr) max_index = arr.index(max_val) left = max(0, max_index-k) right = min(len(arr)-1, max_index+k) selected = arr[left:max_index+1] + arr[max_index:right+1] arr = [x for x in arr if x not in selected] return sum(selected), arr # Input arr = list(map(int, input().split())) a1, b1 = input().split() a2, b2 = input().split() k = int(input()) sum1, sum2 = 0, 0 while arr: s1, arr = get_max_sum(arr, k) sum1 += s1 if not arr: break s2, arr = get_max_sum(arr, k) sum2 += s2 if sum1 > sum2: a1 = toggle(a1) b2 = toggle(b2) else: a2 = toggle(a2) b1 = toggle(b1) xor1 = int(a1, 2) ^ int(b1, 2) xor2 = int(a2, 2) ^ int(b2, 2) if xor1 > xor2: print("Rahul",end="") elif xor2 > xor1: print("Rupesh",end="") else: print("both",end="") Who is Lucky

Pasedd ✅✅✅✅✅✅ Sahre only group everyone ✅✅✅🎉🎉🎉🎉
Pasedd ✅✅✅✅✅✅ Sahre only group everyone ✅✅✅🎉🎉🎉🎉

Join Fast will share One more code soon👇👇👇👇 https://t.me/ibmsolution

Tcs vitaa code ✅✅✅✅✅ All passsd ✅✅✅✅
Tcs vitaa code ✅✅✅✅✅ All passsd ✅✅✅✅

photo content

Join Fast will share One more code soon👇👇👇👇 https://t.me/ibmsolution

Donee tcs code vitaa 1 Share @codeing_area ..
Donee tcs code vitaa 1 Share @codeing_area ..

Sahre group everyone ✅✅✅✅ @codeing_area

X from Y Code C++ Language ✅ Done TCS Codevita #include <bits/stdc++.h> using namespace std; typedef long long ll; #define codeingarea vector<int> #define all(a) (a).begin() , (a).end() #define sz(a) ((int)((a).size())) const ll INF = 4e18; vector<codeingarea> dp(1e4+1 , codeingaarea(1e4+1 , -1)); int f(int i , int j , int s , int r , string &x , string &y , string &y1) {     if(i>j)return 0;     if(dp[i][j] != -1)return dp[i][j];     int mini = 1e9;     for(int k=i; k<=j; k++)     {                 if(y1.find(x.substr(i , k-i+1)) != string::npos)mini = min(mini , r+f(k+1 , j , s , r , x , y , y1));         if(y.find(x.substr(i , k-i+1)) != string::npos)mini = min(mini , s+f(k+1 , j , s , r , x , y , y1));     }     return mini; } int solve(int s , int r , string x, string y) {     string y1 = y;     reverse(all(y1));     return f(0 , sz(x)-1 , s, r , x , y , y1); } int32_t main(){     ios_base::sync_with_stdio(false);      cin.tie(NULL);     string x , y;     cin >> x >> y;     int s, r;     cin >> s >> r;     int ans = solve(s , r, x , y);     cout << ans;     return 0; } X from Y C++ Language 100% Working ✅ TCS Codevita

If you need more answer then Sahre group everyone ❤️✅🎉 @codeing_area

Wait sometime we will share answer soon

If you need more answer then Sahre group everyone ❤️✅🎉 @codeing_area

Tcs code for answer👇 Join 👇👇 https://t.me/ibmsolution