# Perfect power (1, 4, 8, 9, 16, 25, 27, …)

A perfect power is a number that can be expressed as power of another positive integer.

Given a number n, find count of numbers from 1 to n that are of type xy where x >= 1 and y > 1

Examples :

```Input : n = 10
Output : 4
1 4 8 and 9 are the numbers that are
of form x ^ y where x > 0 and y > 1

Input : n = 50
Output : 10
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to go through all powers of numbers from i = 2 to square root of n.

## C++

 `// CPP program to count number of numbers from ` `// 1 to n are of type x^y where x>0 and y>1 ` `#include ` `using` `namespace` `std; ` ` `  `// For our convenience ` `#define ll long long ` ` `  `// Function that keeps all the odd power ` `// numbers upto n ` `int` `powerNumbers(``int` `n) ` `{ ` `    ``// v is going to store all power numbers ` `    ``vector<``int``> v; ` `    ``v.push_back(1); ` ` `  `    ``// Traverse through all base numbers and ` `    ``// compute all their powers smaller than ` `    ``// or equal to n. ` `    ``for` `(ll i = 2; i * i <= n; i++) { ` `        ``ll j = i * i; ` `        ``v.push_back(j); ` `        ``while` `(j * i <= n) { ` `            ``v.push_back(j * i); ` `            ``j = j * i; ` `        ``} ` `    ``} ` ` `  `    ``// Remove all duplicates ` `    ``sort(v.begin(), v.end()); ` `    ``v.erase(unique(v.begin(), v.end()), v.end()); ` ` `  `    ``return` `v.size(); ` `} ` ` `  `int` `main() ` `{ ` `    ``cout << powerNumbers(50); ` `    ``return` `0; ` `} `

## Java

 `// Java program to count number of numbers from ` `// 1 to n are of type x^y where x>0 and y>1 ` `import` `java.io.*; ` `import` `java.util.*; ` ` `  `public` `class` `GFG { ` ` `  `    ``// Function that keeps all the odd power ` `    ``// numbers upto n ` `    ``static` `int` `powerNumbers(``int` `n) ` `    ``{ ` `        ``// v is going to store all unique  ` `        ``// power numbers ` `        ``HashSet v = ``new` `HashSet(); ` `        ``v.add(``1``); ` `      `  `        ``// Traverse through all base numbers ` `        ``// and compute all their powers ` `        ``// smaller than or equal to n. ` `        ``for` `(``int` `i = ``2``; i * i <= n; i++) { ` `            ``int` `j = i * i; ` `            ``v.add(j); ` `            ``while` `(j * i <= n) { ` `                ``v.add(j * i); ` `                ``j = j * i; ` `            ``} ` `        ``} ` `        ``return` `v.size(); ` `    ``} ` `      `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``System.out.print(powerNumbers(``50``)); ` `    ``} ` `} ` `  `  `// This code is contributed by Manish Shaw ` `// (manishshaw1) `

## Python3

 `# Python3 program to count number  ` `# of numbers from 1 to n are of ` `# type x^y where x>0 and y>1 ` ` `  `# Function that keeps all the odd ` `# power numbers upto n ` `def` `powerNumbers(n): ` `     `  `    ``# v is going to store all ` `    ``# unique power numbers ` `    ``v ``=` `set``(); ` `    ``v.add(``1``); ` ` `  `    ``# Traverse through all base ` `    ``# numbers and compute all ` `    ``# their powers smaller than ` `    ``# or equal to n. ` `    ``for` `i ``in` `range``(``2``, n``+``1``): ` `        ``if``(i ``*` `i <``=` `n): ` `            ``j ``=` `i ``*` `i; ` `            ``v.add(j); ` `            ``while` `(j ``*` `i <``=` `n): ` `                ``v.add(j ``*` `i); ` `                ``j ``=` `j ``*` `i; ` ` `  `    ``return` `len``(v); ` `     `  `print` `(powerNumbers(``50``)); ` `# This code is contributed by  ` `# Manish Shaw (manishshaw1) `

## C#

 `// C# program to count number of numbers from ` `// 1 to n are of type x^y where x>0 and y>1 ` `using` `System; ` `using` `System.Collections.Generic; ` `using` `System.Linq; ` ` `  `class` `GFG { ` `     `  `    ``// Function that keeps all the odd power ` `    ``// numbers upto n ` `    ``static` `int` `powerNumbers(``int` `n) ` `    ``{ ` `        ``// v is going to store all unique  ` `        ``// power numbers ` `        ``HashSet<``int``> v = ``new` `HashSet<``int``>(); ` `        ``v.Add(1); ` `     `  `        ``// Traverse through all base numbers ` `        ``// and compute all their powers ` `        ``// smaller than or equal to n. ` `        ``for` `(``int` `i = 2; i * i <= n; i++) { ` `            ``int` `j = i * i; ` `            ``v.Add(j); ` `            ``while` `(j * i <= n) { ` `                ``v.Add(j * i); ` `                ``j = j * i; ` `            ``} ` `        ``} ` `        ``return` `v.Count; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``Console.WriteLine(powerNumbers(50)); ` `    ``} ` `} ` ` `  `// This code is contributed by Manish Shaw ` `// (manishshaw1) `

## PHP

 `0 and y>1 ` ` `  `// Function that keeps all the  ` `// odd power numbers upto n ` `function` `powerNumbers(``\$n``) ` `{ ` `    ``// v is going to store  ` `    ``// all power numbers ` `    ``\$v` `= ``array``(); ` `    ``array_push``(``\$v``, 1); ` ` `  `    ``// Traverse through all base  ` `    ``// numbers and compute all  ` `    ``// their powers smaller than ` `    ``// or equal to n. ` `    ``for` `(``\$i` `= 2; ``\$i` `* ``\$i` `<= ``\$n``; ``\$i``++)  ` `    ``{ ` `        ``\$j` `= ``\$i` `* ``\$i``; ` `        ``array_push``(``\$v``, ``\$j``); ` `        ``while` `(``\$j` `* ``\$i` `<= ``\$n``)  ` `        ``{ ` `            ``array_push``(``\$v``, ``\$j` `* ``\$i``); ` `            ``\$j` `= ``\$j` `* ``\$i``; ` `        ``} ` `    ``} ` ` `  `    ``// Remove all duplicates ` `    ``sort(``\$v``); ` `    ``\$v` `= ``array_unique``(``\$v``); ` ` `  `    ``return` `count``(``\$v``); ` `} ` ` `  `// Driver Code ` `echo` `(powerNumbers(50)); ` ` `  `// This code is contributed by  ` `// Manish Shaw(manishshaw1) ` `?> `

Output:

```10
```

Efficient Solution

We divide output set into subsets.
Even Powers : Simply we need to square root n. The count of even powers smaller than n is square root of n. For example even powers smaller than 25 are (1, 4, 9, 16 and 25).
Odd Powers : We modify above function to consider only odd powers.

## C++

 `// C++ program to count number of numbers from ` `// 1 to n are of type x^y where x>0 and y>1 ` `#include ` `using` `namespace` `std; ` ` `  `// For our convenience ` `#define ll long long ` ` `  `// Function that keeps all the odd power ` `// numbers upto n ` `int` `powerNumbers(``int` `n) ` `{ ` `    ``vector<``int``> v; ` `    ``for` `(ll i = 2; i * i * i <= n; i++) { ` `        ``ll j = i * i; ` `        ``while` `(j * i <= n) { ` `             `  `            ``j *= i; ` ` `  `            ``// We need exclude perfect ` `            ``// squares. ` `            ``ll s = ``sqrt``(j); ` `            ``if` `(s * s != j) ` `                ``v.push_back(j); ` `        ``} ` `    ``} ` ` `  `    ``// sort the vector ` `    ``sort(v.begin(), v.end()); ` `    ``v.erase(unique(v.begin(), v.end()), v.end()); ` ` `  `    ``// Return sum of odd and even powers. ` `    ``return` `v.size() + (ll)``sqrt``(n); ` `} ` ` `  `int` `main() ` `{ ` `    ``cout << powerNumbers(50); ` `    ``return` `0; ` `} `

## Java

 `// Java program to count number  ` `// of numbers from 1 to n are ` `// of type x^y where x>0 and y>1 ` `import` `java.io.*; ` `import` `java.util.*; ` ` `  `class` `GFG ` `{  ` `    ``// Function that keeps all  ` `    ``// the odd power numbers upto n ` `    ``static` `long` `powerNumbers(``int` `n) ` `    ``{ ` `        ``HashSet v = ``new` `HashSet(); ` `        ``for` `(``long` `i = ``2``; i * i * i <= n; i++)  ` `        ``{ ` `            ``long` `j = i * i; ` `            ``while` `(j * i <= n)  ` `            ``{ ` `                ``j *= i; ` `     `  `                ``// We need exclude  ` `                ``// perfect squares. ` `                ``long` `s = (``long``)Math.sqrt(j); ` `                ``if` `(s * s != j) ` `                    ``v.add(j); ` `            ``} ` `        ``} ` `        ``// sort the vector ` `        ``// v.Sort(); ` `        ``// v.erase(unique(v.begin(), ` `        ``// v.end()), v.end()); ` `     `  `        ``// Return sum of odd  ` `        ``// and even powers. ` `        ``return` `v.size() + (``long``)Math.sqrt(n); ` `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``System.out.print(powerNumbers(``50``)); ` `    ``} ` `} ` ` `  `// This code is contributed by  ` `// Manish Shaw(manishshaw1) `

## Python3

 `# Python3 program to count number of  ` `# numbers from 1 to n are of type x^y ` `# where x>0 and y>1 ` `import` `math  ` ` `  `# Function that keeps all the odd power ` `# numbers upto n ` `def` `powerNumbers( n): ` `    ``v ``=` `[] ` `    ``for` `i ``in` `range``(``2``, ` `               ``int``(math.``pow``(n, ``1.0` `/`  `                               ``3.0``)) ``+` `1``) :  ` `        ``j ``=` `i ``*` `i ` `        ``while` `(j ``*` `i <``=` `n) : ` `             `  `            ``j ``=` `j ``*` `i ` ` `  `            ``# We need exclude perfect ` `            ``# squares. ` `            ``s ``=` `int``(math.sqrt(j)) ` `            ``if` `(s ``*` `s !``=` `j): ` `                ``v.append(j) ` `         `  `    ``# sort the vector ` `    ``v.sort()  ` `    ``v ``=` `list``(``dict``.fromkeys(v)) ` ` `  `    ``# Return sum of odd and even powers. ` `    ``return` `len``(v) ``+` `int``(math.sqrt(n)) ` ` `  `# Driver Code ` `if` `__name__``=``=``'__main__'``:  ` `    ``print` `(powerNumbers(``50``)) ` `     `  `# This code is contributed by Arnab Kundu `

## C#

 `// C# program to count number  ` `// of numbers from 1 to n are ` `// of type x^y where x>0 and y>1 ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG ` `{  ` `    ``// Function that keeps all  ` `    ``// the odd power numbers upto n ` `    ``static` `long` `powerNumbers(``int` `n) ` `    ``{ ` `        ``HashSet<``long``> v = ``new` `HashSet<``long``>(); ` `        ``for` `(``long` `i = 2; i * i * i <= n; i++)  ` `        ``{ ` `            ``long` `j = i * i; ` `            ``while` `(j * i <= n)  ` `            ``{ ` `                ``j *= i; ` `     `  `                ``// We need exclude  ` `                ``// perfect squares. ` `                ``long` `s = (``long``)Math.Sqrt(j); ` `                ``if` `(s * s != j) ` `                    ``v.Add(j); ` `            ``} ` `        ``} ` `        ``// sort the vector ` `        ``//v.Sort(); ` `        ``//v.erase(unique(v.begin(), ` `        ``// v.end()), v.end()); ` `     `  `        ``// Return sum of odd  ` `        ``// and even powers. ` `        ``return` `v.Count + (``long``)Math.Sqrt(n); ` `    ``} ` `     `  `    ``// Driver Code ` `    ``static` `void` `Main() ` `    ``{ ` `        ``Console.Write(powerNumbers(50)); ` `    ``} ` `} ` ` `  `// This code is contributed by  ` `// Manish Shaw(manishshaw1) `

## PHP

 `0 and y>1 ` ` `  `// Function that keeps all the ` `// odd power numbers upto n ` `function` `powerNumbers(``\$n``) ` `{ ` `    ``\$v` `= ``array``(); ` `    ``for` `(``\$i` `= 2; ``\$i` `* ``\$i` `* ``\$i` `<= ``\$n``; ``\$i``++)  ` `    ``{ ` `        ``\$j` `= ``\$i` `* ``\$i``; ` `        ``while` `(``\$j` `* ``\$i` `<= ``\$n``)  ` `        ``{ ` `            ``\$j` `*= ``\$i``; ` ` `  `            ``// We need exclude perfect ` `            ``// squares. ` `            ``\$s` `= sqrt(``\$j``); ` `            ``if` `(``\$s` `* ``\$s` `!= ``\$j``) ` `                ``array_push``(``\$v``, ``\$j``); ` `        ``} ` `    ``} ` ` `  `    ``// sort the vector ` `    ``sort(``\$v``); ` `    ``\$uni` `= ``array_unique``(``\$v``); ` `    ``for` `(``\$i` `= 0; ``\$i` `< ``count``(``\$uni``); ``\$i``++) ` `    ``{ ` `        ``\$key` `= ``array_search``(``\$uni``[``\$i``], ``\$v``); ` `        ``unset(``\$v``[``\$key``]); ` `    ``} ` ` `  `    ``// Return sum of odd ` `    ``// and even powers. ` `    ``return` `count``(``\$v``) + 3 +  ` `           ``intval``(sqrt(``\$n``)); ` `} ` ` `  `// Driver Code ` `echo` `(powerNumbers(50)); ` ` `  `// This code is contributed by  ` `// Manish Shaw(manishshaw1) ` `?> `

Output :

```10
```

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