Forwarded from OffCampus Jobs | OnCampus Jobs | Daily Jobs Updates | Lastest Jobs | All Jobs | CSE Jobs | Fresher Jobs โฅ (Dushyant)
Ericsson Hiring Automation Analyst
Link - https://jobs.ericsson.com/careers/job/563121757320992-automation-analyst-gurgaon-haryana-india?domain=ericsson.com
Link - https://jobs.ericsson.com/careers/job/563121757320992-automation-analyst-gurgaon-haryana-india?domain=ericsson.com
Ericsson
Automation Analyst | Gurgaon,Haryana,India | Ericsson
Work with local IT to confirm IT security assessment and data privacy controls (no file can be transmitted without this). Coordinate project timeline with business, receiving vendor, and payroll development to ensure all development, test, and launch datesโฆ
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https://www.linkedin.com/posts/suruchi-dambhare-b0910b171_hexaware-activity-7138860722356207617-OFkN
Linkedin
Suruchi Dambhare on LinkedIn: #hexaware | 144 comments
Hexaware Technologies Ltd. is hiring for Domestic English Voice Process
Job Title: - Customer Support Executive (E-Commerce Domain)
Job Location: - MIHANโฆ | 144 comments on LinkedIn
Job Title: - Customer Support Executive (E-Commerce Domain)
Job Location: - MIHANโฆ | 144 comments on LinkedIn
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Only 2024 Batch
Note : BE/Btech are not eligible
https://usrecruiting.deloitte.com/usi-open-talent-network-risk-and-financial-advisory#subpage/undergraduate/section/nkljb
Note : BE/Btech are not eligible
https://usrecruiting.deloitte.com/usi-open-talent-network-risk-and-financial-advisory#subpage/undergraduate/section/nkljb
Deloitte
Open Talent Network by Risk & Financial Advisory - Deloitte US-India Offices
Deloitte US-India Offices' Risk and Financial Advisory presents Open Talent Network 3.0 - for all final-year students looking for career opportunities.
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Linkedin
Forbes Advisor hiring Intern - Data Research (Looking for 2023 graduates/ BCA/MCA/Bsc-Csc/ BE/ Btech) in Mumbai, Maharashtra, Indiaโฆ
Posted 9:22:09 AM. Company DescriptionForbes Advisor is a new initiative for consumers under the Forbes MarketplaceโฆSee this and similar jobs on LinkedIn.
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Company Name: Autodesk
Batch: 2023 passouts
Role: Software Engineer Apprentice
Apply Link : https://autodesk.wd1.myworkdayjobs.com/en-US/uni/job/Apprentice--Software-Engineer--For-2023-Graduates-Only-_23WD74228
Batch: 2023 passouts
Role: Software Engineer Apprentice
Apply Link : https://autodesk.wd1.myworkdayjobs.com/en-US/uni/job/Apprentice--Software-Engineer--For-2023-Graduates-Only-_23WD74228
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๐ฏ NTrust Infotech Walk-in Drive 2023 for Trainee | 11 to 15 December 2023 ๐ฏ
Job Role : Trainee
Qualification : B.Com/BA/BCA/B.Sc/BBM/BBA
Batch : : 2023
Salary : Rs. 2.15 Lakhs per Annum
Job Role : Trainee
Qualification : B.Com/BA/BCA/B.Sc/BBM/BBA
Batch : : 2023
Salary : Rs. 2.15 Lakhs per Annum
#include <bits/stdc++.h>
using namespace std;
const int MX = 1e6 + 5;
int mu[MX], pre[MX], cnt[MX];
void Sieve() {
vector<int> prime;
vector<bool> isComp(MX, false);
mu[1] = 1;
for (int i = 2; i < MX; ++i) {
if (!isComp[i]) {
prime.push_back(i);
mu[i] = -1;
}
for (int j = 0; j < prime.size() && i * prime[j] < MX; ++j) {
isComp[i * prime[j]] = true;
if (i % prime[j] == 0) {
mu[i * prime[j]] = 0;
break;
} else {
mu[i * prime[j]] = -mu[i];
}
}
}
for (int i = 1; i < MX; ++i) {
pre[i] = pre[i - 1] + mu[i];
}
}
long long KGCD(int N, int K) {
if (K == 0) return 0;
N /= K;
long long res = 0;
for (int l = 1, r; l <= N; l = r + 1) {
r = N / (N / l);
res += 1LL * (N / l) * (N / l) * (pre[r] - pre[l - 1]);
}
return res;
}
int main() {
Sieve();
int T;
cin >> T;
while (T--) {
int N, K;
cin >> N >> K;
cout << KGCD(N, K) << "\n";
}
return 0;
}
GCD Cheggโ
Source :Hola
using namespace std;
const int MX = 1e6 + 5;
int mu[MX], pre[MX], cnt[MX];
void Sieve() {
vector<int> prime;
vector<bool> isComp(MX, false);
mu[1] = 1;
for (int i = 2; i < MX; ++i) {
if (!isComp[i]) {
prime.push_back(i);
mu[i] = -1;
}
for (int j = 0; j < prime.size() && i * prime[j] < MX; ++j) {
isComp[i * prime[j]] = true;
if (i % prime[j] == 0) {
mu[i * prime[j]] = 0;
break;
} else {
mu[i * prime[j]] = -mu[i];
}
}
}
for (int i = 1; i < MX; ++i) {
pre[i] = pre[i - 1] + mu[i];
}
}
long long KGCD(int N, int K) {
if (K == 0) return 0;
N /= K;
long long res = 0;
for (int l = 1, r; l <= N; l = r + 1) {
r = N / (N / l);
res += 1LL * (N / l) * (N / l) * (pre[r] - pre[l - 1]);
}
return res;
}
int main() {
Sieve();
int T;
cin >> T;
while (T--) {
int N, K;
cin >> N >> K;
cout << KGCD(N, K) << "\n";
}
return 0;
}
GCD Cheggโ
Source :Hola
#include<bits/stdc++.h>
using namespace std;
vector<long long> solve(long long m1, long long m2) {
vector<long long> res(3);
for (long long i = 1; ; ++i) {
if (i <= m1 && (i > m2 || m1 >= m2)) {
m1 -= i;
} else if (i <= m2 && (i > m1 || m2 > m1)) {
m2 -= i;
} else {
res[0] = i;
res[1] = m1;
res[2] = m2;
break;
}
}
return res;
}
int main() {
long long m1, m2;
cin >> m1 >> m2;
vector<long long> res = solve(m1, m2);
cout << res[0] << " " << res[1] << " " << res[2] << "\n";
return 0;
}
Computer Crashโ
using namespace std;
vector<long long> solve(long long m1, long long m2) {
vector<long long> res(3);
for (long long i = 1; ; ++i) {
if (i <= m1 && (i > m2 || m1 >= m2)) {
m1 -= i;
} else if (i <= m2 && (i > m1 || m2 > m1)) {
m2 -= i;
} else {
res[0] = i;
res[1] = m1;
res[2] = m2;
break;
}
}
return res;
}
int main() {
long long m1, m2;
cin >> m1 >> m2;
vector<long long> res = solve(m1, m2);
cout << res[0] << " " << res[1] << " " << res[2] << "\n";
return 0;
}
Computer Crashโ
from collections import Counter
def solve(nums, k):
c = Counter(map(lambda x: x.bit_count(), set(nums)))
return sum(c[k1] * c[k2] for k1 in c for k2 in c if k1 + k2 >= k)
t = int(input())
for _ in range(t):
n, k = map(int, input().split())
arr = list(map(int, input().split()))
print (solve(arr, k))
Awesome Pairs โ
def solve(nums, k):
c = Counter(map(lambda x: x.bit_count(), set(nums)))
return sum(c[k1] * c[k2] for k1 in c for k2 in c if k1 + k2 >= k)
t = int(input())
for _ in range(t):
n, k = map(int, input().split())
arr = list(map(int, input().split()))
print (solve(arr, k))
Awesome Pairs โ
#include<bits/stdc++.h>
using namespace std;
int solve(string s, int k) {
int n = s.size();
int count[26] = {0};
int ml = 0;
int mc = 0;
int st = 0;
for (int e = 0; e < n; e++) {
count[s[e] - 'a']++;
mc = max(mc, count[s[e] - 'a']);
if ((e - st + 1) - mc > k) {
count[s[st] - 'a']--;
st++;
}
ml = max(ml, e - st + 1);
}
return ml+2;
}
C++โ
using namespace std;
int solve(string s, int k) {
int n = s.size();
int count[26] = {0};
int ml = 0;
int mc = 0;
int st = 0;
for (int e = 0; e < n; e++) {
count[s[e] - 'a']++;
mc = max(mc, count[s[e] - 'a']);
if ((e - st + 1) - mc > k) {
count[s[st] - 'a']--;
st++;
}
ml = max(ml, e - st + 1);
}
return ml+2;
}
C++โ