Is there any predefined function in c++ to check whether the number is square of any number and same for the cube..

0 votes

0 votes

`sqrt(x)`

, or in general, `pow(x, 1./2)`

or `pow(x, 1./3)`

For example:

```
int n = 9;
int a = (int) sqrt((double) n);
if(a * a == n || (a+1) * (a+1) == n) // in case of an off-by-one float error
cout << "It's a square!\n";
```

Edit: or in general:

```
bool is_nth_power(int a, int n) {
if(n <= 0)
return false;
if(a < 0 && n % 2 == 0)
return false;
a = abs(a);
int b = pow(a, 1. / n);
return pow((double) b, n) == a || pow((double) (b+1), n) == a;
}
```

0 votes

No, but it's easy to write one:

```
bool is_perfect_square(int n) {
if (n < 0)
return false;
int root(round(sqrt(n)));
return n == root * root;
}
bool is_perfect_cube(int n) {
int root(round(cbrt(n)));
return n == root * root * root;
}
```

0 votes

Try this:

```
#include<math.h>
int isperfect(long n)
{
double xp=sqrt((double)n);
if(n==(xp*xp))
return 1;
else
return 0;
}
```

0 votes

No, there are no standard c or c++ functions to check whether an integer is a perfect square or a perfect cube.

If you want it to be fast and avoid using the float/double routines mentioned in most of the answers, then code a binary search using only integers. If you can find an n with n^2 < m < (n+1)^2, then m is not a perfect square. If m is a perfect square, then you'll find an n with n^2=m. The problem is discussed here: https://math.stackexchange.com/questions/41337/efficient-way-to-determine-if-a-number-is-perfect-square

0 votes

For identifying squares i tried this algorithm in java. With little syntax difference you can do it in c++ too. The logic is, the difference between every two consecutive perfect squares goes on increasing by 2. Diff(1,4)=3 , Diff(4,9)=5 , Diff(9,16)= 7 , Diff(16,25)= 9..... goes on. We can use this phenomenon to identify the perfect squares. Java code is,

```
boolean isSquare(int num){
int initdiff = 3;
int squarenum = 1;
boolean flag = false;
boolean square = false;
while(flag != true){
if(squarenum == num){
flag = true;
square = true;
}else{
square = false;
}
if(squarenum > num){
flag = true;
}
squarenum = squarenum + initdiff;
initdiff = initdiff + 2;
}
return square;
}
```

To make the identification of squares faster we can use another phenomenon, the recursive sum of digits of perfect squares is always 1,4,7 or 9. So a much faster code can be...

```
int recursiveSum(int num){
int sum = 0;
while(num != 0){
sum = sum + num%10;
num = num/10;
}
if(sum/10 != 0){
return recursiveSum(sum);
}
else{
return sum;
}
}
boolean isSquare(int num){
int initdiff = 3;
int squarenum = 1;
boolean flag = false;
boolean square = false;
while(flag != true){
if(squarenum == num){
flag = true;
square = true;
}else{
square = false;
}
if(squarenum > num){
flag = true;
}
squarenum = squarenum + initdiff;
initdiff = initdiff + 2;
}
return square;
}
boolean isCompleteSquare(int a){
// System.out.println(recursiveSum(a));
if(recursiveSum(a)==1 || recursiveSum(a)==4 || recursiveSum(a)==7 || recursiveSum(a)==9){
if(isSquare(a)){
return true;
}else{
return false;
}
}else{
return false;
}
}
```

0 votes

For perfect square you can also do:

```
if(sqrt(n)==floor(sqrt(n)))
return true;
else
return false;
```

For perfect cube you can:

```
if(cbrt(n)==floor(cbrt(n)))
return true;
else
return false;
```

Hope this helps.

0 votes

We could use the builtin truc function -

```
#include <math.h>
// For perfect square
bool is_perfect_sq(double n) {
double r = sqrt(n);
return !(r - trunc(r));
}
// For perfect cube
bool is_perfect_cube(double n) {
double r = cbrt(n);
return !(r - trunc(r));
}
```

...