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Forbes Advisor hiring Intern - Data Research (Looking for 2023 graduates/ BCA/MCA/Bsc-Csc/ BE/ Btech) in Mumbai, Maharashtra, Indiaโฆ
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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
vector<int> solve(vector<int>& arr) {
int n = arr[1];
int m = arr[2];
vector<int> img(n * m);
for (int i = 0; i < n * m; i++) {
img[i] = arr[i + 3];
}
vector<int> v(m * n);
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
v[i * n + j] = img[(n - j - 1) * m + i];
}
}
return v;
}
Copy of fix camera issueโ
int n = arr[1];
int m = arr[2];
vector<int> img(n * m);
for (int i = 0; i < n * m; i++) {
img[i] = arr[i + 3];
}
vector<int> v(m * n);
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
v[i * n + j] = img[(n - j - 1) * m + i];
}
}
return v;
}
Copy of fix camera issueโ