8877. Perfect square

A positive integer n is given. If n is a perfect square of some positive integer m, print m. Otherwise, print “No”.

Input. One positive integer n.

Output. Print the required answer.

Sample input 1	Sample output 1
25	5
Sample input 2	Sample output 2
27	No

SOLUTION

mathematics

Algorithm analysis

Let a = . If a * a = n, then n is a perfect square.

Example

Let n = 27. Then a = = 5. Check: 5 * 5 ≠ 27. Therefore, the number 27 is not a perfect square.

Algorithm implementation

Read the input number n.

scanf("%d", &n);

Compute a = .

a = (int)sqrt(n);

If a * a = n, then the number n is a perfect square.

if (a * a == n) printf("%d\n", a);

else puts("No");

Java implementation

import java.util.*;

class Main

{

public static void main(String[] args)

{

Scanner con = new Scanner(System.in);

double n = con.nextDouble();

int a = (int)Math.sqrt(n);

if (a * a == n)

System.out.println(a);

else

System.out.println("No");

con.close();

}

}

Python implementation

Read the input number n.

import math

n = int(input())

Compute a = .

a = int(math.sqrt(n))

If a * a = n, then the number n is a perfect square.

if a * a == n: print(a)

else: print("No")