How to print out perfact squares in given range in C++?

Question

m

Answer 1

A normal square is just something squared - any number, including decimals. The perfect squares are whole integers.

Also, I would just do this:

for (int i = 1; i <= end; ++i) cout << i * i << endl;

It's one line

Answer 2

@Zeke yes but if begin <= i <= end, then
begin <= i*i </= end

Answer 3

//range from begin to end
int i = 0;
while(i * i < begin)
++i;
for(;i * i <= end; ++i)
cout << i*i << endl;

Answer 4

I thought it was the definition of a normal square

Answer 5

👍.this code will help me

Answer 6

void main ( )
{
int begin=1;
int end;
cin >> end;
cout << "perfact squares = \n";
for (;begin*begin<=end;begin++)
{
cout << begin*begin << " ";
}
system ("pause");
}

is this code right?

Answer 7

thanks for help you gyz .this program ran and i got perfact squres in user specified range.bcs i have used your ideas

Answer 8

What is a "perfact square" ? :/ I might be able to help you but I need an explanation on this term for that

Answer 9

A perfect square is a number that can be expressed as the product of two equal integers. E.g. 4, 9, 16, 25, 36, 49, 64, etc...

So use a for loop starting at 2: for (int i = 2; ...
And then output i * i

Answer 10

what would be the syntax for finding perfact squares in given range

Answer 11

yes.thanks