Sir,
my program is giving the right output as per the test cases but still on submitting my code it displays test case #1,#2 : wrong-answer out of the 6 test cases
Compiler Output Issue
#include
#include<math.h>
using namespace std;
int main() {
long long int n,i,y=0;
float x;
cin>>n;
//for(i=0;i<(nn/2 + 1);i++)
while(n>2)
{
++y;
x = pow(yy + nn,0.5);
if((x-(int)x)==0)
{
cout<<y<<" "<<x;
break;
}
}
if(n<=2)//(i==(nn/2 + 1))
{
cout<<"-1";
}
return 0;
}
@sagard0011
Your code is not clear here, can you share in the form of coding blocks ide link and explain your approach?
Also, an alternate approach you can follow is:
If m>n>0 are integers, then (m2-n2, 2mn, m2-n2) is a Pythagorean triple. This is easily seen with a bit of algebra. Thus, plugging in various (m,n) will give various triples.
Now, our goal is to find a triple of the form (m2-n2, 2mn, m2-n2) in which our input value is one of the catheti. Note that the catheti are m2-n2 and 2mn. Also, note that the second cathetus is always even. Thus, it makes sense to consider two cases:
If a is even, then let’s try to equate it with the second cathetus, 2mn, i.e., let’s try to find an (m,n) pair such that 2mn = a. In this case, is even, so we can divide by 2. We have mn = a/2. Thus, all we have to do is to find a factorization of a/2 into mn such that m>n. You can try various factoring methods — even naïve O(√a) time method will pass - but in this case , we don’t really even need to do that, since the trivial (m,n) = (a/2,1) works!. This gives us the solution triple ( (a^2)/4 - 1, a, (a^2)/4 + 1 ). Note that (a^2)/4 is an integer since a is even.
If a is odd, then we can’t equate it with the second cathetus, so let’s try instead m2-n2, i.e. we want to find an (m,n) such that m2-n2 = a. Node that m2-n2 = (m+n)(m-n), so all boils down again to factoring! Let’s try our trivial factorization again: (m+n, m-n) = (a,1) What we get is a system of two linear equations: m+n=a and m-n=1 which has solution m=(a+1)/2 and n=(a-1)/2. Substituting back, this gives us the triple (a, (a^2 - 1)/2, (a^2 + 1)/2). Note that (a^2 + 1)/2 is an integer as a is odd.
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