CODING SOLUTION - Placement Jobs & Materials
前往频道在 Telegram
🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company. “ Contact Admin: Mail me for collab offcampusjobsindiaadmin@gmail.com instagram.com/offcampusjobsindia_it Our Website : www.offcampusjobsindia.com
显示更多📈 Telegram 频道 CODING SOLUTION - Placement Jobs & Materials 的分析概览
频道 CODING SOLUTION - Placement Jobs & Materials (@codingsolution_it) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 129 376 名订阅者，在 职业 类别中位列第 152，并在 印度 地区排名第 1 684 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 129 376 名订阅者。
根据 24 十月, 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 -889，过去 24 小时变化为 -37，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 1.39%。内容发布后 24 小时内通常能获得 0.44% 的反应，占订阅者总量。
- 帖子覆盖： 每篇帖子平均可获得 1 793 次浏览，首日通常累积 565 次浏览。
- 互动与反馈： 受众积极参与，单帖平均反应数为 0。
- 主题关注点： 内容集中在 t.me/codingsolution_it, vector, codevita, string, graduate 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：
“🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company. “
Contact Admin:
Mail me for collab
offcampusjobsindiaadmin@gmail.com
instagram.com/offcampusjobsindia_it
Our Website :
www.offcampusjobsindia.com”
凭借高频更新（最新数据采集于 25 十月, 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 职业 类别中的关键影响点。
129 376
订阅者
-3724 小时
-5037 天
-88930 天
帖子存档
+4
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SELECT
c.name AS NAME,
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SELECT
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#include <bits/stdc++.h>
using namespace std;
struct St {
long long s;
int c;
bool operator<(const St& o) const {
if (s != o.s) return s < o.s;
return c < o.c;
}
};
St mx(St a, St b) {
return a < b ? b : a;
}
St get_mx(const vector<int>& a, long long lam) {
int n = a.size();
St d0 = {0, 0};
St d1 = {-(long long)a[0] - lam, 1};
for (int i = 1; i < n; ++i) {
St n0 = mx(d0, d1);
long long v = -(long long)a[i];
St n1_ext = {d1.s + v, d1.c};
St n1_str = {d0.s + v - lam, d0.c + 1};
St n1 = mx(n1_ext, n1_str);
d0 = n0;
d1 = n1;
}
return mx(d0, d1);
}
long long solve(int n, int k, vector<int>& a) {
long long base = 0;
for (int i = 0; i < n; ++i) {
base += a[i];
}
St r0 = get_mx(a, 0);
long long best = 0;
if (r0.c <= k) {
best = r0.s;
} else {
long long l = 0, r = 2e14;
while (l <= r) {
long long mid = l + (r - l) / 2;
St cur = get_mx(a, mid);
if (cur.c >= k) {
best = cur.s + mid * k;
l = mid + 1;
} else {
r = mid - 1;
}
}
}
return base + 2 * best;
}
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