**Perfect Square**number

**.**

for example: 25 can be expressed as 5 x 5

In the given python3 code we take a user input and take out the square root and convert it into integer. if square root value multiplied by itself equals the user given number then its proved to be a Perfect number else not.

import math n = int(input()) i = int(math.sqrt(n)) #sqrt function returns float so typecasting to int if(n == i*i): print("perfect Square: %d * %d = %d"%(i,i,n)) else: print("Not Perfect Square")

Please comment below if you have a better algorithm than the given one.

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