How To Check Whether A Given Number Is A Perfect Square Or No Check if a given number n is a perfect square or not. if yes then return the number of which it is a perfect square, else print 1. examples: input: n = 4900 output 70 explanation: 4900 is a perfect square number of 70 because 70 * 70 = 4900 input: n = 81 output: 9 explanation: 81 is a perfect square number of 9 because 9 * 9 = 81 approach: to solv. Output 1: enter a number: 121. 121 is a perfect square. output 2: enter a number: 143. 143 is not a perfect square. let's see another logic to check if a given number is perfect square or not. in the following program, we have taken the square root of the number that we have entered and type cast it into int.
Qbasic Program To Check Whether The Input Number Is A Perfect Squar Checking whether number is perfect square or not in c. first of all get the square root of the given number and assign this float value into an integer variable, then only integer part of the number will be stored in integer variable after that compare integer and float variables if their values are same it means number is perfect square. Method 3: using the property that the sum of odd numbers is a perfect square. the given program checks if a number is a perfect square without finding the square root. it does this by iterating over the odd numbers, starting from 1 and subtracting them from the given number n. if n becomes zero, it means that n is a perfect square. time. Program steps. 1. read an integer number from the user. 2. calculate the square root of the given number. 3. check if the square of the square root (when rounded to the nearest integer) equals the original number. 4. display whether the number is a perfect square or not. My answer is: def is square(x): return x**.5 % 1 == 0. it basically does a square root, then modulo by 1 to strip the integer part and if the result is 0 return true otherwise return false. in this case x can be any large number, just not as large as the max float number that python can handle: 1.7976931348623157e 308.
Algorithm And Flowchart To Check Whether A Number Is A Perfect Square Program steps. 1. read an integer number from the user. 2. calculate the square root of the given number. 3. check if the square of the square root (when rounded to the nearest integer) equals the original number. 4. display whether the number is a perfect square or not. My answer is: def is square(x): return x**.5 % 1 == 0. it basically does a square root, then modulo by 1 to strip the integer part and if the result is 0 return true otherwise return false. in this case x can be any large number, just not as large as the max float number that python can handle: 1.7976931348623157e 308. The simplest approach to checking whether a number is a perfect square is to use the math.sqrt () function. if the square root of a number results in an integer, then the number is a perfect square. example: it returns true if the input number is a perfect square (i.e., an integer that is the square of an integer), otherwise false. For every iteration of i, number == i*i will check if the square of i is matching with number. if it matches then the number has a perfect square and we will make the flag=1. if no i value satisfies the perfect square condition, then flag value remains 0. at the end of the program if the flag=0, then the given number is not a perfect square.
Write A Java Program To Check Whether Given Number Is Perfect ођ The simplest approach to checking whether a number is a perfect square is to use the math.sqrt () function. if the square root of a number results in an integer, then the number is a perfect square. example: it returns true if the input number is a perfect square (i.e., an integer that is the square of an integer), otherwise false. For every iteration of i, number == i*i will check if the square of i is matching with number. if it matches then the number has a perfect square and we will make the flag=1. if no i value satisfies the perfect square condition, then flag value remains 0. at the end of the program if the flag=0, then the given number is not a perfect square.
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