Write a C, C++ program to check whether an input number is a perfect square or not. In this program, We will not use an in-built function such as sqrt() to solve this question.
C, C++ programming questions for practice
Find cube of a number
Perfect Square
A number is a perfect square if it is the square of an integer.
For example - 25 is a perfect square. You can write it as 5 * 5 which is the square of an integer. But 17 is not a perfect square.
Output :
Enter a number : 4
4 is a perfect square
Enter a number : 7
7 is not a perfect square
C, C++ programming questions for practice
Find cube of a number
Perfect Square
A number is a perfect square if it is the square of an integer.
For example - 25 is a perfect square. You can write it as 5 * 5 which is the square of an integer. But 17 is not a perfect square.
C Program to Check an Input Number is Perfect Square or not
#include <stdio.h>
int main(void) {
int num, flag = 0;
printf ("Enter a number \n");
scanf ("%d", &num);
for (int i = 1; i <= num; i++) {
if ( i * i == num) {
flag = 1;
break;
}
if ( i * i > num) {
break;
}
}
if (flag) {
printf (" %d is a perfect square", num);
} else {
printf (" %d is not a perfect square", num);
}
return 0;
}
Output :
Enter a number : 4
4 is a perfect square
Enter a number : 7
7 is not a perfect square
C++ Program to Check wether an Input Number is a Perfect Square or not
#include <iostream>
using namespace std;
int main() {
int num, flag = 0;
cout << "Enter a number \n";
cin >> num;
for (int i = 1; i <= num; i++) {
/* If it is equal to an input number
* then it's a perfect square
*/
if ( i * i == num) {
flag = 1;
break;
}
/**
* If it's greater than input number
* so no need to check further
* break the loop
*/
if ( i * i > num) {
break;
}
}
if (flag) {
cout << num << " is a perfect square";
} else {
cout << num << " is not a perfect square";
}
return 0;
}

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