LTM EXAM SOLUTIONS
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π₯Guys plz Stop fearing for daily exams π π¨βπ» @srksvk is here to help you all at lowest cost possible.πͺ π β Our Only Aim Is To Let Get Placed To You In A Reputed Company π₯Effort from our side = π― π±Main Channel: @coding_are π±Tel I'd : @srksvk
Show moreπ Analytical overview of Telegram channel LTM EXAM SOLUTIONS
Channel LTM EXAM SOLUTIONS (@coding_are) in the English language segment is an active participant. Currently, the community unites 14 115 subscribers, ranking 14 118 in the Education category and 28 107 in the India region.
π Audience metrics and dynamics
Since its creation on Π½Π΅Π²ΡΠ΄ΠΎΠΌΠΎ, the project has demonstrated rapid growth, gathering an audience of 14 115 subscribers.
According to the latest data from 03 October, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -103 over the last 30 days and by 2 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 3.70%. Within the first 24 hours after publication, content typically collects 1.68% reactions from the total number of subscribers.
- Post reach: On average, each post receives 522 views. Within the first day, a publication typically gains 237 views.
- Reactions and interaction: The audience actively supports content: the average number of reactions per post is 2.
- Thematic interests: Content is focused on key topics such as placement, gaurntee, suree, capgemini, infosy.
π Description and content policy
The author describes the resource as a platform for expressing subjective opinions:
βπ₯Guys plz Stop fearing for daily exams π
π¨βπ» @srksvk is here to help you all at lowest cost possible.πͺ
π β Our Only Aim Is To Let Get Placed To You In A Reputed Company
π₯Effort from our side = π―
π±Main Channel: @coding_are
π±Tel I'd : @srks...β
Thanks to the high frequency of updates (latest data received on 04 October, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Education category.
14 115
Subscribers
+224 hours
-177 days
-10330 days
Posts Archive
14 105
+2
D.E Shaw exam successfully done by remote access β
β
β
3/3 exam successfully done with all the tets case passed β
π₯
Contact for placement exam @srksvk
14 105
+2
Oracle exam successfully done by π₯π₯π₯π₯
35/35 MCQ successfully done with π― Correct answer β
β
1/1 code successfully executed with all the tets case passed βο»ΏContact for placement exam @srksvk
14 105
+2
Flipkart exam successfully completed π₯π₯π₯π₯π₯
30/30 MCQ done with all 100 π― Correct answer π―π―π―
Contact me placement exam @srksvk
14 105
+1
Lowe's exam successfully done by remote access β
β
β
2/2 codes successfully done with all the tets case passed β
π₯
Contact for placement exam @srksvk
14 105
Oracle exam help available
Contact fast and book your slotss
Contact @srksvk
200% suree clearance guaranteeRemote access also available
14 105
Microsoft exam successfully done by remote access β
β
β
All code and SQL successfully done executed with all tests caes passed β
14 105
Those who want help in ,
Lowes
Accenture,
Oracle
Flipkart
Any off campus or on campus ed exam help available,
Contact @srksvk
200% suree clearance guarantee
Remote access available β
14 105
Those who want help in ,
Lowes
Accenture,
Oracle
Flipkart
Any off campus or on campus ed exam help available,
Contact @srksvk
200% suree clearance guaranteeRemote access available β
14 105
Those who want help in ,
Accenture,
Oracle
Flipkart
Any off campus or on campus ed exam help available,
Contact @srksvk
200% suree clearance guaranteeRemote access available β
14 105
+2
Infosys exam successfully done by remote access β
β
β
β
2 codes fully passed and 1 code half passed β
π₯π₯π₯
Language -java
Contact for placement exam @srksvk
14 105
#include <iostream>
#include <vector>
#include <unordered_map>
#include <algorithm>
using namespace std;
const int MOD = 1e9 + 7;
vector<int> a;
vector<vector<int>> g;
vector<vector<int>> dp;
vector<int> subtree_sum;
int k;
void dfs(int u, int parent) {
dp[u][0] = 1;
subtree_sum[u] = a[u];
for (int v : g[u]) {
if (v != parent) {
dfs(v, u);
vector<int> tmp(k + 1, 0);
for (int i = 0; i <= k; ++i) {
if (dp[u][i]) {
for (int j = 0; j <= k - i; ++j) {
if (dp[v][j]) {
tmp[i + j] = (tmp[i + j] + (long long)dp[u][i] * dp[v][j]) % MOD;
}
}
}
}
for (int i = 0; i <= k; ++i) {
dp[u][i] = (dp[u][i] + tmp[i]) % MOD;
}
}
}
for (int i = k; i >= a[u]; --i) {
dp[u][i] = (dp[u][i] + dp[u][i - a[u]]) % MOD;
}
}
int main() {
int n, m;
cin >> n >> m >> k;
a.resize(n + 1);
g.resize(n + 1);
dp.resize(n + 1, vector<int>(k + 1, 0));
subtree_sum.resize(n + 1, 0);
for (int i = 1; i <= n; ++i) {
cin >> a[i];
}
for (int i = 0; i < m; ++i) {
int x, y;
cin >> x >> y;
g[x].push_back(y);
g[y].push_back(x);
}
long long result = 0;
dfs(1, -1);
for (int i = 1; i <= n; ++i) {
result = (result + dp[i][k]) % MOD;
}
cout << result << endl;
return 0;
}
Tree Function π³π²π΄
14 105
class SegmentTree:
def init(self, data):
self.n = len(data)
self.tree = [0] * (2 * self.n)
self.build(data)
def build(self, data):
for i in range(self.n):
self.tree[self.n + i] = data[i]
for i in range(self.n - 1, 0, -1):
self.tree[i] = max(self.tree[2 * i], self.tree[2 * i + 1])
def range_max(self, left, right):
left += self.n
right += self.n
max_val = 0
while left < right:
if left % 2:
max_val = max(max_val, self.tree[left])
left += 1
if right % 2:
right -= 1
max_val = max(max_val, self.tree[right])
left //= 2
right //= 2
return max_val
def calculate_total_xp(N, A, Bonus):
seg_tree = SegmentTree()
seg_tree.init(Bonus)
last_seen = [-1] * (N // 2 + 1)
right_multiple = [-1] * N
for i in range(N - 1, -1, -1):
power = A[i]
next_pos = float('inf')
for multiple in range(power, N // 2 + 1, power):
if last_seen[multiple] != -1:
next_pos = min(next_pos, last_seen[multiple])
if next_pos != float('inf'):
right_multiple[i] = next_pos
last_seen[power] = i
total_xp = 0
for i in range(N):
if right_multiple[i] != -1:
max_bonus = seg_tree.range_max(i, right_multiple[i] + 1)
total_xp += max_bonus
return total_xp
N = int(input().strip())
A = [int(input().strip()) for _ in range(N)]
Bonus = [int(input().strip()) for _ in range(N)]
result = calculate_total_xp(N, A, Bonus)
print(result)
Some help
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#include <bits/stdc++.h>
using namespace std;
vector<int> sieve(int max_number) {
vector<bool> is_prime(max_number + 1, true);
vector<int> primes;
for (int number = 2; number <= max_number; ++number) {
if (is_prime[number]) {
primes.push_back(number);
for (int multiple = number * number; multiple <= max_number; multiple += number) {
is_prime[multiple] = false;
}
}
}
return primes;
}
int count_divisors(int number, const vector<int>& primes) {
int divisor_count = 0;
int prime_count = primes.size();
for (int subset = 1; subset < (1 << prime_count); ++subset) {
long long least_common_multiple = 1;
int bit_count = 0;
for (int bit = 0; bit < prime_count; ++bit) {
if (subset & (1 << bit)) {
least_common_multiple *= primes[bit];
bit_count++;
if (least_common_multiple > number) break;
}
}
if (least_common_multiple > number) continue;
if (bit_count % 2 == 1) divisor_count += number / least_common_multiple;
else divisor_count -= number / least_common_multiple;
}
return divisor_count;
}
int count_non_divisors(int number, const vector<int>& primes) {
if (number == 0) return 0;
return number - count_divisors(number, primes);
}
int count_non_divisors_range(int max_prime, int left, int right) {
vector<int> primes = sieve(max_prime);
int right_count = count_non_divisors(right, primes);
int left_count = count_non_divisors(left - 1, primes);
return right_count - left_count;
}
int main() {
int max_prime;
string left_str, right_str;
cin >> max_prime >> left_str >> right_str;
cout << count_non_divisors_range(max_prime, stoi(left_str), stoi(right_str)) << endl;
return 0;
}
Divisible string
