Related Articles
Numbers within a range that can be expressed as power of two numbers
• Difficulty Level : Hard
• Last Updated : 24 Apr, 2018

Given two integers L and R. Find the number of perfect powers in the given range [L, R]. A number x is said to be perfect power if there exists some integers a > 0, p > 1 such that x = ap.

Examples :

```Input : 1 4
Output : 2
Explanation :
Suitable numbers are 1 and 4 where 1
can be expressed as 1 = 12
and 4 can be expressed as 4 = 22

Input : 12 29
Output : 3
Explanation :
Suitable numbers are 16, 25 and 27.
```

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

Approach : Let’s fix some power p. It’s obvious that there are no more than 1018/p numbers x such that xp doesn’t exceed 1018 for a particular p. At the same time, only for p = 2 this amount is relatively huge, for all other p ≥ 3 the total amount of such numbers will be of the order of 106. There are 109 squares in the range [1, 1018], so can’t store them to answer our query.

Either, generate all of powers for p ≥ 2 and dispose of all perfect squares among them or generate only odd powers of numbers like 3, 5, 7, etc. Then answer to query (L, R) is equal to the amount of generated numbers between L and R plus some perfect squares in range.

1. The number of perfect squares in the range is the difference of floor value of square root of R and floor value of square root of (L – 1), i.e. (floor(sqrt(R)) – floor(sqrt(L – 1)). Note that due to precision issues the standard sqrt might produce incorrect values, so either use binary search or sqrtl inbuilt function defined in cmath (Check here for more description of sqrtl).
2. To generate those odd powers of numbers. First of all, do precomputation of finding such numbers that can be expressed as power of some number upto 1018 so that we can answer many queries and no need to process them again and again for each query. Start by iterating a loop from 2 to 106 (since we are calculating for powers p ≥ 3 and 106 is the maximum number whose power raised to 3 cannot exceed 1018), for each value we insert its square into a set and check further if that value is already a perfect square (already present in the set), we do not find any other powers of that number (since any power of a perfect square is also a perfect square). Otherwise, run an inside loop to find odd powers of the number until it exceeds 1018 and insert into another set say ‘s’. By this approach, we haven’t pushed any perfect square in the set ‘s’.

Hence the final answer would be sum of number of perfect squares in the range and difference of upper value of R and lower value of L (using binary search).
Below is the implementation of above approach in C++.

 `// CPP Program to count the numbers``// within a range such that number``// can be expressed as power of some``// other number``#include `` ` `using` `namespace` `std;`` ` `#define N 1000005``#define MAX 1e18`` ` `// Vector to store powers greater than 3``vector<``long` `int``> powers;`` ` `// set to store perfect squares``set<``long` `int``> squares;`` ` `// set to store powers other``// than perfect squares``set<``long` `int``> s;`` ` `void` `powersPrecomputation()``{``    ``for` `(``long` `int` `i = 2; i < N; i++) ``    ``{``        ``// pushing squares``        ``squares.insert(i * i);`` ` `        ``// if the values is already``        ``// a perfect square means``        ``// present in the set``        ``if` `(squares.find(i) != squares.end())``                ``continue``;`` ` `        ``long` `int` `temp = i;`` ` `        ``// run loop until some``        ``// power of current number``        ``// doesn't exceed MAX``        ``while` `(i * i <= MAX / temp) ``        ``{``            ``temp *= (i * i);`` ` `            ``/* pushing only odd powers``            ``as even power of a number``            ``can always be expressed as ``            ``a perfect square which is``            ``already present in set squares  */``            ``s.insert(temp);``        ``}``    ``}`` ` `    ``// Inserting those sorted ``    ``// values of set into a vector``    ``for` `(``auto` `x : s)``        ``powers.push_back(x);``}`` ` `long` `int` `calculateAnswer(``long` `int` `L, ``long` `int` `R)``{``    ``// calculate perfect squares in ``    ``// range using sqrtl function``    ``long` `int` `perfectSquares = ``floor``(sqrtl(R)) - ``                            ``floor``(sqrtl(L - 1));`` ` `    ``// calculate upper value of R ``    ``// in vector using binary search``    ``long` `int` `high = (upper_bound(powers.begin(), ``            ``powers.end(), R) - powers.begin());`` ` `    ``// calculate lower value of L ``    ``// in vector using binary search``    ``long` `int` `low = (lower_bound(powers.begin(),``            ``powers.end(), L) - powers.begin());`` ` `    ``// add into final answer``    ``perfectSquares += (high - low);`` ` `    ``return` `perfectSquares;``}`` ` `// Driver Code``int` `main()``{``    ``// precompute the powers``    ``powersPrecomputation();`` ` `    ``// left value of range``    ``long` `int` `L = 12;``     ` `    ``// right value of range``    ``long` `int` `R = 29;`` ` `    ``cout << ``"Number of powers between "` `<< L ``            ``<< ``" and "` `<< R << ``" = "` `<< ``            ``calculateAnswer(L, R) << endl;`` ` `    ``L = 1;``    ``R = 100000000000;`` ` `    ``cout << ``"Number of powers between "` `<< L``            ``<< ``" and "` `<< R << ``" = "` `<< ``            ``calculateAnswer(L, R) << endl;`` ` `    ``return` `0;``}`
Output:
```Number of powers between 12 and 29 = 3
Number of powers between 1 and 100000000000 = 320990
```

My Personal Notes arrow_drop_up