ACCENTURE EXAM SOLUTIONS
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显示更多📈 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 14 139 名订阅者，在 教育 类别中位列第 14 044，并在 印度 地区排名第 28 010 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 14 139 名订阅者。
根据 21 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 -136，过去 24 小时变化为 1，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 4.15%。内容发布后 24 小时内通常能获得 1.53% 的反应，占订阅者总量。
- 帖子覆盖： 每篇帖子平均可获得 586 次浏览，首日通常累积 216 次浏览。
- 互动与反馈： 受众积极参与，单帖平均反应数为 1。
- 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 22 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。
14 139
订阅者
+124 小时
-367 天
-13630 天
帖子存档
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Fully done go and check ✅✅
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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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Infosys 10am exam successfully completed by remote access 👍👍👍👍👍
3/3 hard code done with all tests caes passed ✅✅
Contact for placement exam @srksvk
14 136
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 ✅
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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 ✅
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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
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