Sherlock and Squares


Problem Statement :


Watson likes to challenge Sherlock's math ability. He will provide a starting and ending value that describe a range of integers, inclusive of the endpoints. Sherlock must determine the number of square integers within that range.

Note: A square integer is an integer which is the square of an integer, e.g. .1, 4, 9, 16, 25.

Example
a = 24
b = 49

There are three square integers in the range: 25, 36 and 49. Return 3.


Function Description

Complete the squares function in the editor below. It should return an integer representing the number of square integers in the inclusive range from a to b.

squares has the following parameter(s):

int a: the lower range boundary
int b: the upper range boundary

Returns
int: the number of square integers in the range


Input Format

The first line contains q, the number of test cases.
Each of the next q lines contains two space-separated integers, a and b, the starting and ending integers in the ranges.


Constraints
1 <= q <= 100
1 <= a <= b <= 10^9


Solution :



title-img 


                            Solution in C :

C  :

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main() {

	int i,j;
	int n,t;
	long a,b;
	long c,d;
	//long sqrt;
	 long res = 0;
	scanf("%d",&t);
		
	for(i=0;i<t;i++) {
   	
   	res = 0;
    	 scanf("%d %d",&a,&b);
    	 
    	 c = ceil(sqrt(a));
    	 d = floor(sqrt(b));
    	 
    	 
    	 for(j=c;j<=d;j++) {
    	 	res++;
    	 
    	 }
    	 
    	 printf("%ld\n",res);
  }
	return 0;
}	










C ++  :

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
   int n,m;
  cin>>n;
  
  while (cin>>n>>m)
    {
        cout<<(int)(sqrt(m)+0.0000001)-(int)(sqrt(n-1)+0.0000001)<<endl;
    }
  return 0;
}









Java  :

import java.util.Scanner;
class Solution
{
public static void main(String[] args)
{
	Scanner sc=new Scanner(System.in);
	int t=sc.nextInt();
	for(int h=0;h<t;h++)
	{
	int m=sc.nextInt();
	int n=sc.nextInt();
	int count=0;
	int sq=(int) Math.sqrt(m);
	int a=sq*sq;
        if(a<m)
            {
            sq++;
            a=sq*sq;
        }
     while(a>=m && a<=n)
	{
		count++;
         sq++;
         a=sq*sq;
	
	}
	
	System.out.println(count);
	}
}
}











python 3  :

#!/usr/bin/env python

import sys


if __name__ == '__main__':
    T = int(sys.stdin.readline())
    
    for _ in range(T):
        A, B = list(map(int, sys.stdin.readline().split()))

        nA = int(A**0.5)
        nB = int(B**0.5)
        
        ans = int(B**0.5) - int(A**0.5)
        if nA**2 == A:
        	ans += 1
        
        print(ans)








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