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ACCENTURE EXAM SOLUTIONS

ACCENTURE EXAM SOLUTIONS

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📈 Аналитический обзор Telegram-канала ACCENTURE EXAM SOLUTIONS

Канал ACCENTURE EXAM SOLUTIONS (@coding_are) языкового сегмента Английский является активным участником. Сейчас сообщество объединяет 14 136 подписчиков, занимая 14 103 место в категории Образование и 28 083 место в регионе Индия.

📊 Показатели аудитории и динамика

С момента создания невідомо проект демонстрирует стремительный рост, собрав аудиторию из 14 136 подписчиков.

Согласно последним данным от 21 сентября, 2026, канал показывает стабильную активность. За последние 30 дней изменение числа участников составило -136, а за последние 24 часа — 1, при этом общий охват остаётся высоким.

  • Статус верификации: Не верифицирован
  • Уровень вовлечённости (ER): Средний показатель вовлечённости аудитории составляет 4.15%. В первые 24 часа после публикации контент обычно набирает 1.53% реакций от общего числа подписчиков.
  • Охват публикаций: В среднем каждый пост получает 586 просмотров. В течение первых суток публикация набирает 216 просмотров.
  • Реакции и взаимодействия: Аудитория активно поддерживает контент: среднее количество реакций на один пост — 1.
  • Тематические интересы: Контент сосредоточен на ключевых темах, таких как placement, gaurntee, suree, capgemini, infosy.

📝 Описание и контентная политика

Автор описывает ресурс как площадку для выражения субъективного мнения:
🔥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...

Благодаря высокой частоте обновлений (последние данные получены 23 сентября, 2026) канал поддерживает актуальность и высокий уровень охвата публикаций. Аналитика показывает, что аудитория активно взаимодействует с контентом, что делает его важной точкой влияния в категории Образование.

14 139
Подписчики
+124 часа
-367 дней
-13630 дней
Архив постов
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Taimur and Shadiur Code : ALL PASSED ✅ MOD = 10**9+7 def get_ans(N, Tpos, Shpos): a0 = min(Tpos, Shpos) - 1 b0 = N - max(Tpos, Shpos) dp = [[0]*(b0+1) for _ in range(a0+1)] dp2 = [[0]*(b0+1) for _ in range(a0+1)] dp[a0][b0] = 1 for _ in range(N): for b in range(b0+1): run = 0 for a in range(a0, -1, -1): run = (run + dp[a][b]) % MOD dp2[a][b] = run for a in range(a0+1): run = 0 for b in range(b0, -1, -1): run = (run + dp[a][b]) % MOD dp2[a][b] = (dp2[a][b] + run) % MOD for a in range(a0+1): for b in range(b0+1): dp2[a][b] = (dp2[a][b] - dp[a][b]) % MOD dp, dp2 = dp2, [[0]*(b0+1) for _ in range(a0+1)] return sum(sum(row) for row in dp) % MOD N = int(input()) Tpos = int(input()) Shpos = int(input()) print(get_ans(N, Tpos, Shpos)) *

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int gen_ans(int N, int M, int K, vector<vector<int>> A) { vector<vector<int>> col(N+1, vector<int>(M)); for(int j=0; j<M; j++) for(int i=0; i<N; i++) col[i+1][j] = col[i][j] + (A[i][j] == K); long long Tot = 0, Freq = 0; int MaxB = 0; vector<int> r(M), v(M); for(int i=0; i<N; i++) for(int k=i+1; k<N; k++){ int h = k - i + 1; for(int j=0; j<M; j++){ r[j] = (A[i][j] == K && A[k][j] == K); v[j] = (col[k+1][j] - col[i][j] == h); } for(int j=0; j<M; ){ if(!r[j]) { j++; continue; } int s = j; while(j<M && r[j]) j++; int e = j-1, cnt = 0, f = -1, l = -1; for(int t=s; t<=e; t++) if(v[t]){ if(f<0) f = t; l = t; cnt++; } if(cnt >= 2){ Tot += 1LL*cnt*(cnt-1)/2; int w = l - f + 1; int b = 2*(h + w) - 4; if(b > MaxB){ MaxB = b; Freq = 1; } else if(b == MaxB) Freq++; } } } if(!Tot) return 0; return Tot ^ MaxB ^ Freq; } All passed ✅

def a(s): b = len(s) c = [[0] * b for _ in range(b)] for d in range(b): c[d][d] = 1 for f in range(2, b + 1): for g in range(b - f + 1): h = g + f - 1 if s[g] == s[h] and f == 2: c[g][h] = 2 elif s[g] == s[h]: c[g][h] = c[g + 1][h - 1] + 2 else: c[g][h] = max(c[g][h - 1], c[g + 1][h]) print(c[0][b - 1]) a(input()) https://t.me/coding_are //Infosys LPS - Full pass ✅

One more Done ✅✅ from collections import defaultdict def count(n, c, m, a): mod = 10**9 + 7 start = tuple(a) dp = [defaultdict(int) for _ in range(m+1)] dp[0][start] = 1 for i in range(m): ndp = defaultdict(int) for s in dp[i]: cnt = dp[i][s] ndp[s] = (ndp[s] + cnt) % mod lst = list(s) for j in range(n): if lst[j] == 0: continue if j > 0 and lst[j-1] == 0: new = lst.copy() new[j-1], new[j] = new[j], new[j-1] ndp[tuple(new)] = (ndp[tuple(new)] + cnt) % mod if j < n-1 and lst[j+1] == 0: new = lst.copy() new[j+1], new[j] = new[j], new[j+1] ndp[tuple(new)] = (ndp[tuple(new)] + cnt) % mod dp[i+1] = ndp total = sum(dp[m].values()) % mod return total Language pyhton ✅

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