finding the number of perfect squares in the given range - c++

Question

I am trying to find out the number of perfect squares in the given range. The method I am following is provided on: digital_root
The code I have implemented is not giving correct answers sometimes because This method does not consider the numbers like 10,1000 etc.
Please help me to work out this method.
int cot=0;
void squares(int a,int b){
if(a==b){
int digit,digit_root=0,no;
no=a;
digit=no%10;
if(digit==2||digit==3||digit==7||digit==8){
}else{
no=a;
while(no>0){
digit=no%10;
if((digit==0)||(digit==9)){
}else{
digit_root=digit_root+digit;
digit_root=digit_root%9;
}
no=no/10;
}
if(digit_root==0||digit_root==7||digit_root==1||digit_root==4){
if(digit_root==1){
if(a)
}
cot++;
}
}
}else{
int c=(a+b)/2;
squares(a,c);
squares(c+1,b);
}
}
int main() {
int a,b,t;
cin>>t;
for(int i=0;i<t;i++){
cin >> a>>b;
squares(a,b);
cout << cot<<endl;
cot=0;
}
return 0;
}

Answer 1

the best algorithm will probably be to subtract sqrt(bigger number) - sqrt(other num), i.e. something like this
int a = 1,b=100;
if (sqrt(a) != int(sqrt(a)){
cout<<int(sqrt(b))-int(sqrt(a))<<endl;}
else{
cout<<int(sqrt(b))-int(sqrt(a))+1<<endl;}
the logic is very simple, perfect square is natural num * the same natural num so, 1*1,2*2,3*3,4*4 and so on
So you just need to get the closest perfect sqrt of the bigger num and subtract it with the smaller num (just be careful in case that the smallest num is a perfect square itself)

Answer 2

There's too much going on here. Simplify it. Write a function that determines whether a single number is a perfect square. Get that working right. Then use that function to determine how many perfect squares there are in your range. That probably shouldn't be recursive; that's just too hard to follow. Use a loop that covers the range:
for (int i = a; i <= b; ++i)
if (is_perfect_square(i))
++count;

Answer 3

It's easy to print numbers from 1 to N, and the numbers will be perfect square.
Here is the logic to check perfect square
/*function definition to check perfect square*/
int isPerfectSquare(int number)
{
int iVar;
float fVar;
fVar=sqrt((double)number);
iVar=fVar;
if(iVar==fVar)
return 1;
else
return 0;
}

Answer 4

You dont need all these costly calculations.
Things are really simple here.
Instead of looping from A to B and checking for perfect squares,
we can calculate the first integer square root of A at O(1) and
count the rest perfect squares with simple increments of power n,
up to B. In this way we can avoid costly calculations.
But even better, we can even avoid loops by just using ceil and floor.
For example, square numbers between 1000 and 2000:
sqrt(1000) = 31.6 --> ceil(31.6) --> 32
sqrt(2000) = 44.7 --> floor(44.7)--> 44
therefore the difference of these two plus one gives us the number of perfect squares.
#include <cmath>
#include <iostream>
using namespace std;
int main() {
int A, B;
//get the range [A,B]
cin >> A >> B;
cout << floor(sqrt(B)) - ceil(sqrt(A)) + 1 << end;
return 0;
}