Determining the number of square integers (perfect squares) between any given range.

/*Square integer is a integer which is a perfect square of any integer. In this , I have tried this with 2 possible solutions*/

1st solution (brute force method):-

#include<stdio.h>
#include<math.h>
int main(){
    int n;
    long t,a,b;
    int i,count=0;
    printf("Enter number of test cases\n");//number of ranges
    scanf("%d",&n);
    while(n>0){
        count=0;
        scanf("%d%d",&a,&b);
        for(i=a;i<=b;i++){
            t=sqrt(i);/*function to determine square root of any given number;*/
           
            if(t*t==i)/*checking if this particular number is perfect square*/

            count++;
           
        }
        n--;
    printf("%d\n",count);
    }
}

2nd solution (one of the best(optimized way) to determine if number is perfect square or not):-

#include<stdio.h>
#include<math.h>
int main(){
    int n,a,b;
    int count=0;
    printf("Enter number of test cases\n");//number of ranges
    scanf("%d",&n);
    while(n>0){
        scanf("%d%d",&a,&b);
        count=floor(sqrt(b))-ceil(sqrt(a))+1;/*optimized way to calculate all perfect squares b/w a &&b;*/
 
        printf("%d\n",count);
        n--;
    }

o/p:-
Enter number of test cases
2
3 9
2 /*between 3 and 9 (inclusive) only 2 numbers are perfect squares--4 and 9*/

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