Sir, I am not able to think a way to limit my time please help me

#1

code is :

#include
using namespace std;

int main(){

unsigned long long int n;
cin >> n;
unsigned long long int x = n;
x++;
unsigned long long int y = x;
y++;

while(((n*n) + (x*x)) != y*y){
	
	while(((n*n) + (x*x)) > y*y){
		y++;
	}
	while(((n*n) + (x*x)) < y*y){
		x++;
	}
}
cout << x << " " << y;

}

#2

Hey @prerakvarshney12
First, suppose n is the shortest side of the triangle, m, k are other two sides. According to Pythagorean Theorem, we know n^2 + m ^2 = k ^2

just do a change k^2 - m ^2 = n ^2

futherly ( k + m )( k - m ) = n^2

as we know, n ^2 × 1 = n ^2 so we can suppose that k + m = n ^2, k - m = 1

easily, we get

because the side is a interger, so this solution can only be used when n is a odd.

So how to deal with even? we find that if (k-m) is odd, the solution is suitable for odd. On the other hand, we guess that if(k-m) is even, the solution is suitable for even.

just as this,

so, we get

this is an O (1) algorithm.

#3

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