ACCENTURE EXAM SOLUTIONS
前往频道在 Telegram
🔥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
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 14 138 名订阅者，在 教育 类别中位列第 14 046，并在 印度 地区排名第 28 032 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 14 138 名订阅者。
根据 20 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 -145，过去 24 小时变化为 -3，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 4.14%。内容发布后 24 小时内通常能获得 1.54% 的反应，占订阅者总量。
- 帖子覆盖： 每篇帖子平均可获得 585 次浏览，首日通常累积 217 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 21 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。
14 137
订阅者
-324 小时
-347 天
-14530 天
帖子存档
14 137
+1
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Any off campus or on campus exam help available
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public static List getMinSum(List arr, List query, int k) {
List result = new ArrayList<>();
int n = arr.size();
for (int z : query) {
int minSum = Integer.MAX_VALUE;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
int x = arr.get(i);
int y = arr.get(j);
if (x == 0 && y == 0) continue;
if (x != 0 && z % x == 0) {
int a = z / x, b = 0;
if (a >= 0) minSum = Math.min(minSum, a + b);
}
if (y != 0 && z % y == 0) {
int a = 0, b = z / y;
if (b >= 0) minSum = Math.min(minSum, a + b);
}
if (x != 0) {
for (int a = 0; a <= k; a++) {
int rem = z - a * x;
if (rem < 0) break;
if (y != 0 && rem % y == 0) {
int b = rem / y;
if (b >= 0 && a + b <= k) {
minSum = Math.min(minSum, a + b);
}
}
}
}
}
}
result.add(minSum <= k ? minSum : -1);
}
return result;
}
14 137
import java.util.*;
class Result {
public static int findMinOperations(List<Integer> arr) {
int n = arr.size();
int operations = 0;
int rootCount = 0;
for (int x : arr) {
if (x == -1) rootCount++;
}
if (rootCount == 0) {
arr.set(0, -1);
operations++;
} else if (rootCount > 1) {
boolean found = false;
for (int i = 0; i < n; i++) {
if (arr.get(i) == -1) {
if (!found) {
found = true;
} else {
arr.set(i, 1);
operations++;
}
}
}
}
for (int i = 0; i < n; i++) {
int parent = arr.get(i);
if (parent != -1 && (parent < 1 || parent > n)) {
arr.set(i, 1);
operations++;
}
}
boolean[] visited = new boolean[n];
boolean[] inStack = new boolean[n];
for (int i = 0; i < n; i++) {
if (!visited[i]) {
operations += dfsFixCycle(arr, visited, inStack, i);
}
}
return operations;
}
private static int dfsFixCycle(List<Integer> arr, boolean[] visited, boolean[] inStack, int node) {
int operations = 0;
while (node != -1 && !visited[node]) {
visited[node] = true;
inStack[node] = true;
int parent = arr.get(node) == -1 ? -1 : arr.get(node) - 1;
if (parent != -1) {
if (!visited[parent]) {
node = parent;
} else if (inStack[parent]) {
arr.set(node, -1);
operations++;
break;
} else {
break;
}
} else {
break;
}
}
Arrays.fill(inStack, false);
return operations;
}
}
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
List<Integer> arr = new ArrayList<>();
for (int i = 0; i < n; i++) {
arr.add(sc.nextInt());
}
System.out.println(Result.findMinOperations(arr));
}
}
14 137
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