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๐ MetricStream Off Campus Recruitment Drive 2021 | BE/ B.Tech | MCA | Freshers | Bangalore
* Job Role : Graduate Engineer Trainee-Member Technical Staff
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โ ๐ช @fresherearth โ
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โ ๐ช @fresherearth โ
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Capgemini
Network Trainee
@f_a_a_n_g_777
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@sup777exams
2020 & 2019 batch
code:- 86817371
Network Trainee
@f_a_a_n_g_777
https://app.talentrank.in/a/rt/1683
@sup777exams
2020 & 2019 batch
code:- 86817371
Capgemini
Network Engineer
@f_a_a_n_g_777
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@sup777exams
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Code:-36397349
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@f_a_a_n_g_777
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@f_a_a_n_g_777
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@sup777exams
careers.mobiveil.com
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๐ฏCODENATION OFF CAMPUS DRIVE
Role: Software Development Engineer
Eligibility: B.E/B.Tech, MCA
Batch:2021
CTC: 29.5 Lakh Per Annum
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Telegram: @fresherearth
Role: Software Development Engineer
Eligibility: B.E/B.Tech, MCA
Batch:2021
CTC: 29.5 Lakh Per Annum
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Telegram: @fresherearth
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๐ Hexaview Off Campus Drive 2021 | Freshers | Application Engineer | BE/ B.Tech/ ME/ M.Tech | Pune/ Noida | 4-6 LPA
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โ ๐ช @fresherearth โ
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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
int solve (int n, int m, vector<int> v) {
long long l = 1, r = *max_element(v.begin(), v.end()), mid, cnt;
while(l<r){
mid = (l+r)/2;
cnt = 0;
for(auto& i: v) cnt += ceil(1.0*i/mid);
if(cnt>n) l = mid+1;
else r = mid;
}
return l;
}
Seimens vaccination distribution code
#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
int solve (int n, int m, vector<int> v) {
long long l = 1, r = *max_element(v.begin(), v.end()), mid, cnt;
while(l<r){
mid = (l+r)/2;
cnt = 0;
for(auto& i: v) cnt += ceil(1.0*i/mid);
if(cnt>n) l = mid+1;
else r = mid;
}
return l;
}
Seimens vaccination distribution code
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For checking number is power of two or not
Python
Python