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<!DOCTYPE html>
<html>
<head>
<title>Polaris</title>
<link rel="stylesheet" type="text/css" href="bootstrap/css/bootstrap.css">
<link rel="stylesheet" type="text/css" href="bootstrap/css/bootstrap.min.css">
<link rel="stylesheet" type="text/css" href="index.css">
</head>
<body>
<nav class="navbar mt-2">
<h1 class="mx-5">Polaris</h1>
<div>
<a href="">Home</a>
<a href="">Home</a>
<a href="">Home</a>
<a href="">Home</a>
</div>
</nav>
<div class="container mt-5">
<div class="row">
<div class="col-8 mt-5">
<h1>Welcome to Polaris</h1>
</div>
<div class="col-4 mt-1">
<h2>Get in touch with us</h2>
<form class="form">
<input type="email" name="email" class="form-control mt-5" placeholder="Email">
<input type="text" name="fullname" class="form-control mt-3" placeholder="Full Name">
<input type="number" name="phoneNumber" class="form-control mt-3" placeholder="Business Phone">
<input type="text" name="company" class="form-control mt-3" placeholder="Company Name">
<textarea cols="20" rows="5" class="form-control mt-3" placeholder="How can we help you?"></textarea>
<center>
<input id="submit" type="submit" value="submit" class="mt-5 btn">
</center>
</form>
</div>
</div>
</div>
</html>
Telegram:::: @placementupdatess
Siemens coding answer
#include<stdio.h>
#include<string.h>
#include<map>
#include<algorithm>
#define ll long long
using namespace std;
map<ll,ll>mp[30][30];
int n,m,t;
ll k,num[30][30],ans;
// double-ended dfs, one from (1,1) to (i,j) and i+j==max(n,m)
void dfs1(int x,int y,ll tt)
{
if(x+y==t)
{
Mp[x][y][tt]++; //Use the map to record the number of tt when searching for points (y, x)
return ;
}
if(x+1<=n)
dfs1(x+1,y,tt^num[x+1][y]);
if(y+1<=m)
dfs1(x,y+1,tt^num[x][y+1]);
}
// second from (n,m) to (i,j) and (i+j)==max(n,m)+1
void dfs2(int x,int y,ll tt)
{
if(x+y==t+1)
{
if(x-1>=1)
ans+=mp[x-1][y][tt^k];
if(y-1>=1)
ans+=mp[x][y-1][tt^k];
return ;
}
if(x-1>=1)
dfs2(x-1,y,tt^num[x-1][y]);
if(y-1>=1)
dfs2(x,y-1,tt^num[x][y-1]);
}
int main()
{
while(~scanf("%d%d%lld",&n,&m,&k))
{
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
mp[i][j].clear();
scanf("%lld",&num[i][j]);
}
}
ans=0;
t=max(n,m);
if(1+1<=t)
{
dfs1(1,1,num[1][1]);
dfs2(n,m,num[n][m]);
}
else
{
if(num[1][1]==k)
ans++;
}
printf("%lld\n",ans);
}
}
XoR paths
Juspay Solutions:
@placementupdatess
--1st
def traverse(graph, visited, curr, path, stack, n):
if curr==n-1:
path.append(stack.copy())
stack.pop()
return path, stack
else:
visited[curr] = 1
for ele in graph[curr]:
if visited[ele] == 0:
stack.append(ele)
path, stack = traverse(graph, visited, ele, path, stack, n)
stack.pop()
return path, stack
n, m, t, c = map(int, input().split())
graph = [[] for _ in range(n)]
visited = [0 for _ in range(n)]
for _ in range(m):
a,b = map(int, input().split())
graph[a-1].append(b-1)
graph[b-1].append(a-1)
path, _ = traverse(graph, visited, 0, [], [0], n)
path.sort(key=lambda x:len(x))
# print(path)
tim = 0
l = len(path[0])
for i in range(1, l):
tim += c
if i==l-1:
break
else:
temp = tim//t
if temp%2==1:
tim = (temp+1)*t
print(tim)
2nd--
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
vector<int> store[1001];
vector<pair<ll,vector<ll>>> ptr;
void dfs(ll st,ll e,ll vis[],vector<ll> rs,ll w){
rs.push_back(st);
if(st == e){
ptr.push_back({w*(rs.size()-1),rs});
return;
}
for(auto u : store[st]){
if(vis[u] == 0){
vis[st] = 1;
dfs(u,e,vis,rs,w);
vis[st] = 0;
}
}
}
int main()
{
ll n,m,t,c,u,v;
cin>>n>>m>>t>>c;
while(m--){
cin>>u>>v;
store[u].push_back(v);
store[v].push_back(u);
}
vector<ll> rs;
ll w = c;
ll vis[n+1] = {0};
dfs(1,n,vis,rs,w);
sort(ptr.begin(),ptr.end());
vector<ll> rt[n+1];
for(int i=0;i<ptr.size();i++){
ll nes = ptr[i].second.size();
for(auto u : ptr[i].second){
rt[u].push_back(nes);
}
}
ll trt[n+1] = {0};
trt[1] = 1;
trt[n] = 1;
for(int i=2;i<=n-1;i++){
if(rt[i].size() > 0){
ll tm = rt[i][0];
ll up = upper_bound(rt[i].begin(),rt[i].end(),tm) - rt[i].begin();
trt[i] = up;
}
}
for(int i=1;i<=n;i++)
cout<<trt[i]<<" ";
return 0;
}
3rd--
def fun(arr,s,en,length,char,t,ans,final,green,ch):
if ans//t>ch:
green=not green
ch+=1
if s==en:
final.append(ans)
return
for i in arr[s]:
if str(i) not in length[:-1].split():
if green:
fun(arr,i,en,length+str(i)+" ",char,t,ans+char,final,green,ch)
else:
aq=ans%t
fun(arr,i,en,length+str(i)+" ",char,t,ans+char+t-aq,final,not green,ch)
return
n,m,t,c=map(int,input().split())
d={}
for i in range(m):
u,v=map(int,input().split())
if u in d:
d[u].append(v)
else:
d[u]=[v]
if v in d:
d[v].append(u)
else:
d[v]=[u]
l="1 "
final=[]
green=True
ch=0
ret=fun(d,1,n,l,c,t,0,final,green,ch)
final=set(final)
final=list(final)
final.sort()
if len(final)>1:
print(final[1])
else:
print(-1)
Telegram : @placementupdatess
from math import ceil
for _ in range(int(input())):
n,d=map(int,input().split())
l=list(map(int,input().split()))
r=0
for i in range(len(l)):
if(l[i]<=9 or l[i]>=80):
r+=1
print(ceil(r/d)+ceil((n-r)/d))
#Vaccine distribution
LG Soft Off Campus Drive | 2019 2020 & 2021 Batch | 3.4LPAbit.ly/35SWSOL Telegram - t.me/placementupdatess
Swpna regular is a reader
https://docs.google.com/forms/d/e/1FAIpQLScsJrXZya9qPVz4AUbXcDVxFxhAXgQKq62ZIYzkvTP2pzgQIA/viewform
Google Docs
GTA Recruitment'21
Global Technology Alliance (GTA) is an order engineering team working in collaboration within Schneider electric manufacturing plants in North America
Company : SHARDA MOTOR INDUSTRIES
Job Role: ENGINEER – BUSINESS SOLUTION
Location : Chennai
Qualifications : B.Tech/BE in Mechanical
Experience : 0 - 2 Years
Apply Link : https://cutt.ly/PmwKzee
Faiz, [25.06.21 22:37]
Need 50 Mechanical diploma candidates with or without experience for BOSCH
Pune location.
Pls inform anybody if u know.
Salary 35 to 45 k (take home).
2 years bond. Confirmation based on work efficiency after 2 years.
Immediate requirement.
praveen.ks@in.bosch.com
If it is not useful to you...
Share this in your circle, this create someone's career.
Mechanical budda buddis can apply.🏃♂️🏃♂️🏃♂️🏃♂️🏃♂️🏃♂️
Note others are not eligible🚫
