2025 año en números

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Suscriptores
-3624 horas
-3367 días
-1 20130 días
Archivo de publicaciones
Company : NTT
Role : Intern
Qualification :Bachelor’s / Master’s
Batch : 2025 & 2026
Salary : Up to ₹ 6LPA
Experience : Fresher’s
Location : Bengaluru
Apply link : https://careers.services.global.ntt/global/en/job/NTT1GLOBALR121946EXTERNALENGLOBAL/Intern?utm_source=indeed&utm_medium=phenom-feeds
Company : Capgemini
Role : Deal Centre of Excellence
Qualification : Graduate/B Com/BAF
Batch : 2024 without any Active Backlogs
Salary : Up to ₹ 5 LPA
Experience : Fresher’s
Location : Mumbai
Apply link : https://www.capgemini.com/in-en/solutions/off-campus-drive-for-deal-centre-of-excellence-dcoe-2024-graduates/
using System;
using System.Collections.Generic;
class Program
{
static void Main()
{
int N = int.Parse(Console.ReadLine());
var graph = new Dictionary>();
var indegrees = new Dictionary();
for (int i = 0; i < N; i++)
{
var edge = Console.ReadLine().Split();
string from = edge[0];
string to = edge[1];
if (!graph.ContainsKey(from))
graph[from] = new List();
graph[from].Add(to);
if (!indegrees.ContainsKey(from))
indegrees[from] = 0;
if (!indegrees.ContainsKey(to))
indegrees[to] = 0;
indegrees[to]++;
}
var words = Console.ReadLine().Split();
var levels = new Dictionary();
var queue = new Queue();
foreach (var node in indegrees.Keys)
{
if (indegrees[node] == 0)
{
levels[node] = 1; // Root level is 1
queue.Enqueue(node);
}
}
while (queue.Count > 0)
{
var current = queue.Dequeue();
int currentLevel = levels[current];
if (graph.ContainsKey(current))
{
foreach (var neighbor in graph[current])
{
if (!levels.ContainsKey(neighbor))
{
levels[neighbor] = currentLevel + 1;
queue.Enqueue(neighbor);
}
indegrees[neighbor]--;
if (indegrees[neighbor] == 0 && !levels.ContainsKey(neighbor))
{
queue.Enqueue(neighbor);
}
}
}
}
int totalValue = 0;
foreach (var word in words)
{
if (levels.ContainsKey(word))
{
totalValue += levels[word];
}
else
{
totalValue += -1;
}
}
Console.Write(totalValue);
}
}
String Puzzle
C+
👍 2
int main() {
string s;
cin >> s;
int n = s.length(), res = 0;
vector v(n);
for (int i = 0; i < n; ++i) cin >> v[i];
int lw = s[0] - '0', lwv = v[0];
for (int i = 1; i < n; ++i) {
if (s[i] - '0' == lw) {
res += min(lwv, v[i]);
lwv = max(lwv, v[i]);
} else {
lw = s[i] - '0';
lwv = v[i];
}
}
cout << res;
return 0;
}
Form alternating string
Change accordingly before submitting to avoid plagiarism
#include<stdio.h>
int main()
{
int n;
scanf("%d",&n);
int arr[n];
for(int i=0;i<n;i++)
{
scanf("%d",&arr[i]);
}
int count=0;
int num=arr[0];
for(int i=1;i<n;i++)
{
if(num!=arr[i])
count++;
}
printf("%d",count);
}
Count press
int main() {
string s;
cin >> s;
int n = s.length(), res = 0;
vector v(n);
for (int i = 0; i < n; ++i) cin >> v[i];
int lw = s[0] - '0', lwv = v[0];
for (int i = 1; i < n; ++i) {
if (s[i] - '0' == lw) {
res += min(lwv, v[i]);
lwv = max(lwv, v[i]);
} else {
lw = s[i] - '0';
lwv = v[i];
}
}
cout << res;
return 0;
}
#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <cmath>
#include <algorithm>
using namespace std;
struct Line {
int x1, y1, x2, y2;
};
int countCells(Line line, pair<int, int> star, bool split) {
if (line.x1 == line.x2) {
if (split) {
return min(abs(star.second - line.y1), abs(star.second - line.y2)) + 1;
}
else {
return abs(line.y1 - line.y2) + 1;
}
}
else {
if (split) {
return min(abs(star.first - line.x1), abs(star.first - line.x2)) + 1;
}
else {
return abs(line.x1 - line.x2) + 1;
}
}
}
bool intersects(Line a, Line b, pair<int, int>& intersection) {
if (a.x1 == a.x2 && b.y1 == b.y2) {
if (b.x1 <= a.x1 && a.x1 <= b.x2 && a.y1 <= b.y1 && b.y1 <= a.y2) {
intersection = {a.x1, b.y1};
return true;
}
}
if (a.y1 == a.y2 && b.x1 == b.x2) {
if (a.x1 <= b.x1 && b.x1 <= a.x2 && b.y1 <= a.y1 && a.y1 <= b.y2) {
intersection = {b.x1, a.y1};
return true;
}
}
return false;
}
int main() {
int N, K;
cin >> N;
vector<Line> lines(N);
for (int i = 0; i < N; ++i) {
cin >> lines[i].x1 >> lines[i].y1 >> lines[i].x2 >> lines[i].y2;
if (lines[i].x1 > lines[i].x2 || (lines[i].x1 == lines[i].x2 && lines[i].y1 > lines[i].y2)) {
swap(lines[i].x1, lines[i].x2);
swap(lines[i].y1, lines[i].y2);
}
}
cin >> K;
map<pair<int, int>, vector<Line>> stars;
for (int i = 0; i < N; ++i) {
for (int j = i + 1; j < N; ++j) {
pair<int, int> intersection;
if (intersects(lines[i], lines[j], intersection)) {
stars[intersection].push_back(lines[i]);
stars[intersection].push_back(lines[j]);
}
}
}
int asiylam = 0;
for (auto& star : stars) {
if (star.second.size() / 2 == K) {
vector<int> intensities;
for (auto& line : star.second) {
intensities.push_back(countCells(line, star.first, true));
}
asiylam += *min_element(intensities.begin(), intensities.end());
}
}
cout << asiylam << endl;
return 0;
}
Magic Star Intensity Code
C++
TCS Exam
int main() {
int n, m, k, d = 1, rv = 0;
cin >> n >> m;
vector> f(n);
for (int i = 0, u, v; i < m; ++i) {
cin >> u >> v;
f[u].insert(v);
f[v].insert(u);
}
cin >> k;
vector w(n, true);
rv = n;
while (rv < k) {
vector nw(n, false);
for (int i = 0; i < n; ++i) {
int cnt = 0;
for (int fr : f[i]) cnt += w[fr];
if (w[i] && cnt == 3) nw[i] = true;
else if (!w[i] && cnt < 3) nw[i] = true;
}
w = nw;
rv += count(w.begin(), w.end(), true);
++d;
}
cout << d;
return 0;
}
Office Rostering
C+
TCS exam
👍 1
#include<stdio.h>
int main()
{
int n;
scanf("%d",&n);
int arr[n];
for(int i=0;i<n;i++)
{
scanf("%d",&arr[i]);
}
int count=0;
int num=arr[0];
for(int i=1;i<n;i++)
{
if(num!=arr[i])
count++;
}
printf("%d",count);
}
C Language
TCS 1st Qsn
---------------------------------------------------------
N=int(input())
K=int(input())
price=list(map(int,input().split()))
vol=list(map(int,input().split()))
maxvol=0
volu=0
maxvol=max(vol)
for i in range(0,N):
if (maxvol==vol[i] and price[i]<=K):
K=K-price[i]
volu=maxvol
for i in range(0,N):
for j in range(i+1,N+1):
if (price[i]<=K and price[i]==price[j]):
if (vol[i]>vol[j]):
volu=volu+vol[i]
K=K-price[i]
else:
volu=volu+vol[j]
K=K-price[j]
elif (price[i]<=K and price[i]!=price[j]):
K=K-price[i]
-------
include<stdio.h>
int main()
{
int n;
scanf("%d",&n);
int arr[n];
for(int i=0;i<n;i++)
{
scanf("%d",&arr[i]);
}
int count=0;
int num=arr[0];
for(int i=1;i<n;i++)
{
if(num!=arr[i])
count++;
}
printf("%d",count);
}
Array Code in C language
👍 1
Capstoneco Hiring
Summer Internship
Batch: 2025/2026
Location: London, New York
Duration: 10 Weeks
Apply now:- https://www.itjobs.services/2024/11/capstoneco-hiring.html
👍 3
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👍 4
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MIUM RESOURCES
➙Data science
➙Python
➙Artificial Intelligence .
➙AWS Certified .
➙Cloud
➙BIG DATA
➙Data Analytics
➙BI
➙Google Cloud Platform
➙IT Training
➙MBA
➙Machine Learning
➙Deep Learning
➙Ethical Hacking
➙SPSS
➙Statistics
➙Data Base
➙Learning language resources ( English , French , German )
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1. CAPGEMINI is hiring fresher candidates for the role of Off-campus. The details of the job, requirements and other information given below: https://www.capgemini.com/in-en/solutions/off-campus-drive-for-deal-centre-of-excellence-dcoe-2024-graduates/
2. DELOITTE is hiring fresher candidates for the role of USI |ESC| Associate Analyst. The details of the job, requirements and other information given below: https://usijobs.deloitte.com/careersUSI/JobDetail/FY25-USI-Audit-Services-EH-ESC-Associate-Analyst/194295
3. TECH MAHINDRA is hiring fresher candidates for the role of Customer Support Executive. The details of the job, requirements and other information given below: https://www.naukri.com/job-listings-customer-support-executive-tech-mahindra-hyderabad-0-to-5-years-231124005738?src=jobsearchDesk&sid=1732363626886947_8&xp=2&px=2
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👍 6❤ 1
Adobe is hiring for Software Development Engineer Role
*Role: Software Development Engineer
*Location:Noida, IN
*Category: Software Engineering
Employment Type: Full-time
Apply Now:- www.itjobs.services
👍 4❤ 2
