Hackerrank's Problem Solving solutions( Divisible Sum Pairs, Migratory Birds, Day of the Programmer ) :
Problem 16: Divisible Sum Pairs
Solution: (in c++)
( please guys before moving to the solution try it yourself at least 3-4 times , if you really wanna become a good coder)
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n,k;
cin>>n>>k;
int a[n];
for(int p=0;p<n;p++)
{
cin>>a[p];
}
int count=0;
for(int i=0; i<n; i++)
{
for(int j=i+1; j<n; j++)
{
if((a[i]+a[j])%k==0)
{
count++;
}
}
}
cout<<count;
}
This code is simple. There's no need for any explanation.
Problem 17: Migratory Birds
Solution: (in c++)
( please guys before moving to the solution try it yourself at least 3-4 times , if you really wanna become a good coder)
#include<bits/stdc++.h>
using namespace std;
int main()
{
map<long long ,long long>m; //declaring a map
long long n,i,max=0,p;
cin>>n;
long long a[n];
for(i=0;i<n;i++)
{
cin>>a[i];
}
for(i=0;i<n;i++)
{
m[a[i]]++; //you will understand this once you have done map STL
}
for(auto &e : m) //map traversal
{
if(max<e.second)
{
max=e.second;
p=e.first;
}
}
cout<<p;
}
This code is simple. There's no need for any explanation.
Problem 18: Day of the Programmer
Solution: (in c++)
( please guys before moving to the solution try it yourself at least 3-4 times , if you really wanna become a good coder)
#include<bits/stdc++.h>
using namespace std;
int leap(int n)
{
if(n==1918)
{
return 2;
}
else if(n>=1919) //---leap year program----
{
if(n%100==0 && n%400==0)
{
return 1;
}
else
{
if(n%4==0 && n%100!=0)
return 1;
else
return 0;
}
} //----------------------
else //i.e (n<1918)
{
if(n%4==0)
{
return 1;
}
else
{
return 0;
}
}
}
int main()
{
int y,p;
cin>>y;
p=leap(y);
if(p==1)
{
cout<<"12.09."<<y;
}
else if(p==0)
{
cout<<"13.09."<<y;
}
else
{
cout<<"26.09."<<y;
}
}
This code is simple. There's no need for any explanation.
Guys , if you have any queries or need more explanation for something , comment below!
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