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ACCENTURE EXAM SOLUTIONS

Kanalga Telegram’da o‘tish

🔥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

Ko'proq ko'rsatish

Hindiston28 164Taʼlim14 646...

📈 Telegram kanali ACCENTURE EXAM SOLUTIONS analitikasi

ACCENTURE EXAM SOLUTIONS (@coding_are) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 13 949 obunachidan iborat bo'lib, Taʼlim toifasida 14 646-o'rinni va Hindiston mintaqasida 28 164-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 13 949 obunachiga ega bo‘ldi.

03 Noyabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -151 ga, so‘nggi 24 soatda esa -6 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

Tasdiqlash holati: Tasdiqlanmagan
Jalb etish (ER): Auditoriya o‘rtacha 3.70% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.38% ini tashkil etuvchi reaksiyalarni to‘playdi.
Post qamrovi: Har bir post o‘rtacha 516 marta ko‘riladi; birinchi sutkada odatda 193 ta ko‘rish yig‘iladi.
Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 2 ta reaksiya keladi.
Tematik yo‘nalishlar: Kontent placement, gaurntee, suree, capgemini, infosy kabi asosiy mavzularga jamlangan.

📝 Tavsif va kontent siyosati

Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
“🔥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...”

Yuqori yangilanish chastotasi (oxirgi ma’lumot 04 Noyabr, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Taʼlim toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.

13 949

Obunachilar

-624 soatlar

-377 kun

-15130 kun

516

Post ko'rishlar

~ 19324 soatlar

~ 26148 soatlar

3.70%

Muloqot nisbati

~ 3

Kuniga postlar

Ads index

Postlar arxiv

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In c language

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$#include #define MAX_SIZE 100 int main() {     char str[MAX_SIZE];     char toSearch;  &nb$

#include <stdio.h> #define MAX_SIZE 100 int main() { char str[MAX_SIZE]; char toSearch; int i, count; gets(str); toSearch = getchar(); count = 0; i=0; while(str[i] != '\0') { if(str[i] == toSearch) { count++; } i++; } printf("%d",count); return 0; } Tech Mahindra

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#include <bits/stdc++.h> using namespace std; #include <bits/stdc++.h> using namespace std; int main() { int n; cin >> n; vector<int> inputArr, answer; for(int i=0; i<n; ++i) { int temp; cin >> temp; inputArr.push_back(temp); } for(int i=0; i<n; ++i) if(inputArr[i]%2==0) answer.push_back(inputArr[i]); for(int i=0; i<n; ++i) if(inputArr[i]%2!=0) answer.push_back(inputArr[i]); for(int i=0; i<n; ++i) cout << answer[i] << " "; return 0; } 2nd code -C++ langugage

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#include <stdio.h> int main() { int num,tNum,digit,cnt; int rem; scanf("%d",&num); scanf("%d",&digit); cnt=0; tNum=num; while(tNum>0) { rem=tNum%10; if(rem==digit) cnt++; tNum/=10; } printf("%d",cnt); return 0; } Tech Mahindra first code

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Tech Mahindra help done💯💯💯💥💥

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tcs nqt free help to all if i have 500 members

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https://t.me/rituyr

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Please share the group everyone. In this group you will get help for free coding questions

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def solution(a,b): res=a.count(b) if res>0: return res else: return 0 a=input() b=input() print(solution(a,b))

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