# PROBLEM LINK: Contest Practice

**Author:** Baban Gain

**Editorialist:** Baban Gain

# DIFFICULTY:

EASY

# PREREQUISITES:

Basic Mathematics

# PROBLEM:

For a given number N, You need to find no. of perfect squares or perfect cubes from 1 to N ( both inclusive )

Note : If a number, like 64, which is both square ( 8^2 ) and cube (4^3) should be counted once only.

# EXPLANATION:

Brute force solutions should fail as the question has tight time limit.

The answer is very simple and needs no explanation

ans = sqrt(N) + cbrt(N) - sqrt(cbrt(N)) for every test case.

where sqrt represents Square Root

cbrt represents Cube Root

sqrt(N) represents No. of squares upto N, cbrt(N) represents No. of cubes upto N.

But there can be numbers that can be both perfect square and perfect cube, which are considered twice.

Hence we can eliminate extra counting by subtracting sqrt(cbrt(N))

# ALTERNATIVE SOLUTION:

ans = sqrt(N) + cbrt(N) - cbrt(sqrt(N))

# AUTHORâ€™S SOLUTIONS:

Authorâ€™s solution can be found here.