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allcoding1_official

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📈 Telegram 频道 allcoding1_official 的分析概览

频道 allcoding1_official (@allcoding1_official) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 82 159 名订阅者,在 技术与应用 类别中位列第 1 505,并在 印度 地区排名第 3 684

📊 受众指标与增长动态

невідомо 创建以来,项目保持高速增长,吸引了 82 159 名订阅者。

根据 27 八月, 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 -1 527,过去 24 小时变化为 -53,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 3.53%。内容发布后 24 小时内通常能获得 0.72% 的反应,占订阅者总量。
  • 帖子覆盖: 每篇帖子平均可获得 2 899 次浏览,首日通常累积 588 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 dsa, stack, namaste, javascript, dev 等核心主题上。

📝 描述与内容策略

尚未提供频道描述。

凭借高频更新(最新数据采集于 28 八月, 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 技术与应用 类别中的关键影响点。

82 159
订阅者
-5324 小时
-2957
-1 52730
帖子存档
int solve(int N, int days, vector& weights) { int start_cap = 1; int end_cap = 0; for (int cargo_w : weights) { start_cap = max(start_cap, cargo_w); end_cap += cargo_w; } end_cap *= 2; int min_required_capacity = end_cap; while (start_cap <= end_cap) { int mid_cap = start_cap + (end_cap - start_cap) / 2; int active_day = 1; int loaded_weight = 0; int pkg_idx = 0; bool fits_in_time = true; while (pkg_idx < N) { if (active_day > days) { fits_in_time = false; break; } int day_capacity = (active_day % 2 == 1) ? mid_cap : (mid_cap / 2); if (weights[pkg_idx] > day_capacity) { active_day++; loaded_weight = 0; continue; } if (loaded_weight + weights[pkg_idx] <= day_capacity) { loaded_weight += weights[pkg_idx]; pkg_idx++; } else { active_day++; loaded_weight = 0; } } if (fits_in_time && active_day <= days) { min_required_capacity = mid_cap; end_cap = mid_cap - 1; } else { start_cap = mid_cap + 1; } } return min_required_capacity; }

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int solve(int N, int P, vector>& costs) { int half = N / 2; const int MAX_VAL = 1e9; vector>> table(N + 1, vector>(half + 1, vector(2, MAX_VAL))); table[1][1][0] = costs[0][0]; table[1][0][1] = costs[0][1]; for (int idx = 1; idx < N; idx++) { for (int cityA_cnt = 0; cityA_cnt <= half; cityA_cnt++) { for (int prev_city = 0; prev_city < 2; prev_city++) { if (table[idx][cityA_cnt][prev_city] == MAX_VAL) continue; if (cityA_cnt + 1 <= half) { int extra = (prev_city == 0) ? P : 0; table[idx + 1][cityA_cnt + 1][0] = min(table[idx + 1][cityA_cnt + 1][0], table[idx][cityA_cnt][prev_city] + costs[idx][0] + extra); } int cityB_cnt = idx - cityA_cnt; if (cityB_cnt + 1 <= half) { int extra = (prev_city == 1) ? P : 0; table[idx + 1][cityA_cnt][1] = min(table[idx + 1][cityA_cnt][1], table[idx][cityA_cnt][prev_city] + costs[idx][1] + extra); } } } } return min(table[N][half][0], table[N][half][1]); }

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long long ans = 0; for (int i = 0; i &lt; N; i++) { for (int j = i + 1; j &lt; N; j++) { int k = 2 * j - i; if (k &gt;= N) br
long long ans = 0; for (int i = 0; i < N; i++) { for (int j = i + 1; j < N; j++) { int k = 2 * j - i; if (k >= N) break; long long a = nums[i]; long long b = nums[j]; long long c = nums[k]; long long x = min({a, b, c}); long long z = max({a, b, c}); long long y = a + b + c - x - z; if (x + y > z) ans = max(ans, a + b + c); } }

Infosys Exam 2PM BE READY I WILL SHARE ANSWERS Loading..... Please share with your friends @allcoding1_official

Infosys Exam 2PM BE READY I WILL SHARE ANSWERS Please share with your friends

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import java.util.*; class Main { public static long solve(int N, int M, int G, int[] A, int[] B) { if (N == 0 || M == 0) return 0; int maxL = 1; for (int offset = -(M - 1); offset < N; offset++) { int startA = Math.max(0, offset); int endA = Math.min(N - 1, offset + M - 1); int len = endA - startA + 1; if (len <= maxL) continue; int[] diffs = new int[len]; for (int k = 0; k < len; k++) { int idxA = startA + k; int idxB = idxA - offset; diffs[k] = A[idxA] - B[idxB]; } Map> shiftMap = new HashMap<>(); for (int k = 0; k < len; k++) { shiftMap.computeIfAbsent(diffs[k], x -> new ArrayList<>()).add(k); } for (int d : shiftMap.keySet()) { List exactIndices = shiftMap.get(d); int nExact = exactIndices.size(); for (int e = 0; e < nExact; e++) { int left = exactIndices.get(e); int right = exactIndices.get(e); int graceCount = 0; int curLen = 0; for (int k = left; k < len; k++) { if (diffs[k] == d) { curLen++; } else if (Math.abs(diffs[k] - d) <= G && graceCount == 0) { graceCount++; curLen++; } else { break; } } for (int k = left - 1; k >= 0; k--) { if (diffs[k] == d) { curLen++; } else if (Math.abs(diffs[k] - d) <= G && graceCount == 0) { graceCount++; curLen++; } else { break; } } maxL = Math.max(maxL, curLen); } } } return maxL; }

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int solve(int n, int index, int maxSum) { auto getSum = [](long long peak, long long count, long long diff) { long long k = m
int solve(int n, int index, int maxSum) { auto getSum = [](long long peak, long long count, long long diff) { long long k = min(count, (peak - 1) / diff); long long sum = k * peak - diff * k * (k + 1) / 2; sum += (count - k); return sum; }; auto possible = [&](long long peak) { long long left = getSum(peak, index, 1); long long right = getSum(peak, n - index - 1, 2); long long total = peak + left + right; return total <= maxSum; }; long long low = 1; long long high = maxSum; long long ans = 1; while (low <= high) { long long mid = low + (high - low) / 2; if (possible(mid)) { ans = mid; low = mid + 1; } else { high = mid - 1; } } return (int)ans; }

import java.util.*; class Main { static final int MOD = 1000000007; public static int solve(int N, int F, int T) { int m = N - 2; // dp[s] = ways to obtain sum s using processed nodes long[] dp = new long[T + 1]; dp[0] = 1; for (int i = 0; i < m; i++) { long[] ndp = new long[T + 1]; for (int sum = 0; sum <= T; sum++) { if (dp[sum] == 0) continue; for (int v = 1; v <= F && sum + v <= T; v++) { ndp[sum + v] += dp[sum]; if (ndp[sum + v] >= MOD) ndp[sum + v] %= MOD; } } dp = ndp; } long waysPerPosition = 0; for (int a = 1; a <= F; a++) { for (int b = 1; b <= F; b++) { int need = T - a * b; if (need >= 0 && need <= T) { waysPerPosition += dp[need]; waysPerPosition %= MOD; } } } return (int) ((waysPerPosition * (N - 1)) % MOD); }

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import java.util.*; class Main { public static int solve(int N, int[] arr) { int maxVal = 0; for (int x : arr) { if (x &gt; m
import java.util.*; class Main { public static int solve(int N, int[] arr) { int maxVal = 0; for (int x : arr) { if (x > maxVal) maxVal = x; } int[] freq = new int[maxVal + 1]; for (int x : arr) { freq[x]++; } int need = N / 2; int left = 1; int count = 0; int ans = Integer.MAX_VALUE; for (int right = 1; right <= maxVal; right++) { count += freq[right]; while (count >= need) { ans = Math.min(ans, right - left + 1); count -= freq[left]; left++; } } return ans; }

#include <bits/stdc++.h> using namespace std; long long sideSum(long long v, long long count, long long step) { if (count <= 0) return 0; long long full = min(count, (v - 1) / step); long long sum = full * v - step * full * (full + 1) / 2; sum += (count - full) * 1LL; return sum; } bool feasible(long long v, long long n, long long index, long long maxSum) { long long left = index; long long right = n - 1 - index; long long total = v + sideSum(v, left, 1) + sideSum(v, right, 2); return total <= maxSum; } int solve(int n, int index, int maxSum) { long long lo = 1, hi = maxSum, ans = 1; while (lo <= hi) { long long mid = lo + (hi - lo) / 2; if (feasible(mid, n, index, maxSum)) { ans = mid; lo = mid + 1; } else { hi = mid - 1; } } return (int)ans; } int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); int n; cin >> n; int index; cin >> index; int maxSum; cin >>

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import java.util.*; class Main { public static int solve(int P, int M, int k, int[][] piles) { List best = new ArrayList<>(); for (int i = 0; i < P; i++) { ArrayList a = new ArrayList<>(); for (int j = 0; j < M; j++) { if (piles[i][j] != -1) a.add(piles[i][j]); } int n = a.size(); int[] mx = new int[n + 1]; Arrays.fill(mx, Integer.MIN_VALUE); mx[0] = 0; if (n > 0) { int[] pre = new int[n + 1]; for (int j = 0; j < n; j++) pre[j + 1] = pre[j] + a.get(j); for (int len = 1; len <= n; len++) { int bestSum = Integer.MIN_VALUE; for (int s = 0; s + len <= n; s++) { bestSum = Math.max(bestSum, pre[s + len] - pre[s]); } mx[len] = bestSum; } } best.add(mx); } int NEG = -1000000000; int[] dp = new int[k + 1]; Arrays.fill(dp, NEG); dp[0] = 0; for (int[] cur : best) { int[] ndp = new int[k + 1]; Arrays.fill(ndp, NEG); int lim = cur.length - 1; for (int used = 0; used <= k; used++) { if (dp[used] == NEG) continue; for (int take = 0; take <= lim && used + take <= k; take++) { ndp[used + take] = Math.max( ndp[used + take], dp[used] + cur[take] ); } } dp = ndp; } return dp[k]; }

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👉Ashok IT AWS Devops Azure Devops 👉Naresh IT All courses 50 rupees Lifetime access Contact:- @meterials_available