I need to know if the sum of two numbers is a perfect square, I thought of adding the two and taking the root and then checking if the rest of the division by 1 is greater than zero, except that the operator (%) only works with integers and sqrt does not return me int, someone has a better idea or knows how can I build my own MOD function?
if(sqrt(a+b)%1==0) ...
One possibility is to do a binary search. So you do not need to use float.
unsigned intSqrt(int N){
// invariante: lo*lo <= N < hi*hi
unsigned lo = 0;
unsigned hi = 1<<16;
while(hi - lo > 1){
unsigned mid = (lo + hi)/2;
unsigned mid2 = mid*mid;
if( mid2 < N ){
lo = mid;
}else if(mid2 == N){
return mid;
}else{ // N < mid2
hi = mid;
}
}
return lo;
}
int quadradoPerfeito(unsigned N){
int x = intSqrt(N);
return x*x == N;
}
int quadperfeito(int n)
{
return (sqrt(n)-((int)sqrt(n)) == 0);
}
int somaEhQuadperfeito(int m, int n)
{
return quadperfeito(m+n);
}
I found a solution:
int raiz_int = sqrt(a+b);
double raiz_double = sqrt(a+b);
if (raiz_int == raiz_double) ... ;
else ...
Not the best way but