Pythagorean triplet

#1

#include
#include<math.h>
using namespace std;
int main()
{
int n;
cin>>n;
float h,b;
if(n%2==0)
{
h=pow((n/2),2)+1;
b=pow((n/2),2)-1;
if(h==int(h) && b==int(b))
{
cout<<b<<endl;
cout<<h<<endl;
}
else
{
cout<<"-1"<<endl;
}
}
else if(n%2!=0)
{
h=(pow(n,2)+1)/2;
b=(pow(n,2)-1)/2;
if(h==int(h) && b==int(b))
{
cout<<b<<endl;
cout<<h<<endl;
}
else
{
cout<<"-1"<<endl;
}
}
return 0;
}

What is the problem in this code?

#2

is the input provided not the smallest side of the pythagorean triplet??

#3

hey @dsdishu99 answer is -1 only when n < 3 else the answer exist so you dont need to check in other cases.
and since n can be as large as 10^9 n squared can be as large as 10^18 so dont use pow() function just do n * n and use long long and double. Here is code for your reference

#4

#include #include<math.h> using namespace std; int main() { int n; cin>>n; long long int h,b; while(n>3) { if(n%2==0) { h=((n/2)(n/2))+1; b=((n/2)(n/2))-1; if(h==int(h) && b==int(b)) { cout<<b<<endl; cout<<h<<endl; } else { cout<<"-1"<<endl; } } else if(n%2!=0) { h=((nn)+1)/2; b=((nn)-1)/2; if(h==int(h) && b==int(b)) { cout<<b<<endl; cout<<h<<endl; } else { cout<<"-1"<<endl; } } } return 0; }

#5

still not passig all test cases… :frowning:

#6

@dsdishu99 please share your code through, ide.codingbocks.com, it is very difficult to read and work with code in this way.

#7

#include #include<math.h> using namespace std; int main() { int n; cin>>n; long long int h,b; while(n>3) { if(n%2==0) { h=((n/2)(n/2))+1; b=((n/2)(n/2))-1; if(h==int(h) && b==int(b)) { cout<<b<<endl; cout<<h<<endl; } else { cout<<"-1"<<endl; } } else if(n%2!=0) { h=((nn)+1)/2; b=((nn)-1)/2; if(h==int(h) && b==int(b)) { cout<<b<<endl; cout<<h<<endl; } else { cout<<"-1"<<endl; } } } return 0; }

#8
  1. open ide.codingblocks.com
  2. paste your code their
  3. click on save
  4. copy the link and paste it here
#9
#10

@dsdishu99
Use break statements when you print anything
Some sample cases to check :

Q 23
A 264 265

Q 1
A -1

Q 99999979
A 4999997900000220 4999997900000221

#11


please check…not producing output as per second and third instance you suggested…

#12

@dsdishu99
This is the solution

If your doubt is resolved please mark it as closed.

#13

is it not in c language while i am learniing c++…?

#14

please tell me where am i lagging in my solutin?

#15

@dsdishu99
This was C++ only. C++ can also use C based input output and header files style.
Here is your modified solution https://ide.codingblocks.com/s/192415

If your doubt is resolved please mark it as closed.

#16

how would it conclude if it even exists or not…as given in the problem we gotta print -1 when triplet doesn’t exist??

#17

@dsdishu99
For n>3 triplet always exists

#18

so -1 will be returned only if n<3??

#19

@dsdishu99
Yes thats right

#20

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