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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Sherlock and Squares
  5. Discussions

Sherlock and Squares

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1593 Discussions

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  • labesangela
     + 0 comments

    a1=math.ceil(math.sqrt(a)) b1=math.floor(math.sqrt(b)) roots=(x for x in range(a1,b1+1)) roots1=list(roots) return len(roots1)

  • e2248473
     + 0 comments
    Here is my Java Code =============== 
    public static int squares(int a, int b) {
    boolean status = Math.sqrt(a)% 1 ==0;
    int startRoot = (int) Math.sqrt(a);
    int endRoot = (int) Math.sqrt(b);if (status){
    return endRoot-startRoot+1;
}else{
    return endRoot-startRoot;
}


}

  • yashm27729
     + 0 comments

    Python3 solution:

    def squares(a, b):
    # Write your code here
    count = 0
    x = math.sqrt(a)
    if x == int(x):
        rangex = x
    else:
        rangex = int(math.ceil(x))
        
    y = math.sqrt(b)
    if y == int(y):
        rangey = y
    else:
        rangey = int(math.floor(y))
        
    if rangey < rangex:
        return count
        
    for i in range(int(rangex),int(rangey+1)):
        count += 1
    
    return count

  • vu_241fa04549
     + 0 comments

    for _ in range(int(input())): import math a,b=map(int,input().split()) print(math.floor(b0.5) - math.ceil(a0.5) + 1)

    This is solution for begginer

  • banibrataghosh91
     + 0 comments
    def squares(a, b):
    # Write your code here
    count = 0
    start = int( math.sqrt(a))
    end = int(math.sqrt(b))
    
    for i in range(start,end+1):
        if i**2 >=a and i**2<=b:
            count+=1
            
    return count