Endi mavjud! Telegram Tadqiqoti 2025 โ yilning asosiy insaytlari 
MTHREE exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer
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๐ฅGuys plz Stop fearing for daily exams ๐ ๐จโ๐ป @srksvk is here to help you all at lowest cost possible.๐ช ๐ โ Our Only Aim Is To Let Get Placed To You In A Reputed Company ๐ฅEffort from our side = ๐ฏ ๐ฑMain Channel: @coding_are ๐ฑTel I'd : @srksvk
Ko'proq ko'rsatish๐ Telegram kanali MTHREE exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer analitikasi
MTHREE exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer (@coding_are) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 13 236 obunachidan iborat bo'lib, Taสผlim toifasida 15 345-o'rinni va Hindiston mintaqasida 32 011-o'rinni egallagan.
๐ Auditoriya koโrsatkichlari va dinamika
ะฝะตะฒัะดะพะผะพ sanasidan buyon loyiha tez oโsib, 13 236 obunachiga ega boโldi.
19 Iyun, 2026 dagi oxirgi maโlumotlarga koโra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -137 ga, soโnggi 24 soatda esa -4 ga oโzgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya oโrtacha 2.81% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.07% ini tashkil etuvchi reaksiyalarni toโplaydi.
- Post qamrovi: Har bir post oโrtacha 372 marta koโriladi; birinchi sutkada odatda 142 ta koโrish yigโiladi.
- Reaksiyalar va oโzaro taโsir: Auditoriya faol: har bir postga oโrtacha 2 ta reaksiya keladi.
- Tematik yoโnalishlar: Kontent placement, gaurntee, suree, capgemini, infosy kabi asosiy mavzularga jamlangan.
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Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida taโriflaydi:
โ๐ฅGuys plz Stop fearing for daily exams ๐
๐จโ๐ป @srksvk is here to help you all at lowest cost possible.๐ช
๐ โ Our Only Aim Is To Let Get Placed To You In A Reputed Company
๐ฅEffort from our side = ๐ฏ
๐ฑMain Channel: @coding_are
๐ฑTel I'd : @srks...โ
Yuqori yangilanish chastotasi (oxirgi maโlumot 20 Iyun, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli boโlib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Taสผlim toifasidagi muhim taโsir nuqtasiga aylantirishini koโrsatadi.
13 236
Obunachilar
-424 soatlar
-407 kunlar
-13730 kunlar
Postlar arxiv
+1
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#include<bits/stdc++.h>
using namespace std;
int k, r, m, res = 0;
vector<string> words;
void solve(int i, vector<int>& lan, int cnt, const vector<string>& qw) {
ย if (i == (int)qw.size()) {
ย ย ย res = max(res, cnt);
ย ย ย return;
ย }
ย if (cnt + (int)(qw.size() - i) <= res) return;
ย for (int j = 0; j < r; j++) {
ย ย ย if (lan[j] == 0) {
ย ย ย ย ย lan[j] = qw[i].size();
ย ย ย ย ย solve(i + 1, lan, cnt + 1, qw);
ย ย ย ย ย lan[j] = 0;
ย ย ย ย ย break;
ย ย ย } else if (lan[j] + 1 + (int)qw[i].size() <= m) {
ย ย ย ย ย lan[j] += 1 + qw[i].size();
ย ย ย ย ย solve(i + 1, lan, cnt + 1, qw);
ย ย ย ย ย lan[j] -= 1 + qw[i].size();
ย ย ย }
ย }
ย solve(i + 1, lan, cnt, qw);
}
int main() {
ย ios::sync_with_stdio(false);
ย cin.tie(0);
ย cin >> k;
ย words.resize(k);
ย for (auto& s : words) cin >> s;
ย cin >> r >> m;
ย vector<string> qw;
ย for (auto& s : words) if ((int)s.size() <= m) qw.push_back(s);
ย sort(qw.begin(), qw.end(), [&](const string& a, const string& b) -> bool {
ย ย ย if (a.size() != b.size()) return a.size() > b.size();
ย ย ย return a < b;
ย });
ย vector<int> lan(r, 0);
ย solve(0, lan, 0, qw);
ย cout << res;
}
faulty segment
Equation Puzzle Code
C++
TCS
int main() {
string e1, e2;
int r;
cin >> e1 >> e2 >> r;
// @codeing_are
int a1, b1, c1, d1, n1, a2, b2, c2, d2, n2;
sscanf(e1.c_str(), "%dx+%dy+%dz+%dw<=%d", &a1, &b1, &c1, &d1, &n1);
sscanf(e2.c_str(), "%dx+%dy+%dz+%dw<=%d", &a2, &b2, &c2, &d2, &n2);
vector<vector<int>> combinations;
for (int x = 0; x <= r; x++) {
for (int y = 0; y + x <= r; y++) {
for (int z = 0; z + x + y <= r; z++) {
int w = r - x - y - z;
combinations.push_back({x, y, z, w});
}
}
}
set<int> results;
for (auto &vars1 : combinations) {
int v1 = a1 * vars1[0] + b1 * vars1[1] + c1 * vars1[2] + d1 * vars1[3];
if (v1 > n1) continue;
for (auto &vars2 : combinations) {
int v2 = a2 * vars2[0] + b2 * vars2[1] + c2 * vars2[2] + d2 * vars2[3];
if (v2 > n2) continue;
if (v1 == v2) results.insert(v1);
}
}
cout << results.size() << endl;
return 0;
}
Equation Puzzle Code
C++
TCS
