Count Square Root, Cube Root and Both in Given Range using C++

Suppose we have a number n. We have to count the number of integers x, in range 1 to n such that, that x is a square of a positive integer number or a cube of a positive integer number (or both a square and a cube simultaneously).

Problem Category

Various problems in programming can be solved through different techniques. To solve a problem, we have to devise an algorithm first, and to do that we have to study the particular problem in detail. A recursive approach can be used if there is a recurring appearance of the same problem over and over again; alternatively, we can use iterative structures also. Control statements such as if-else and switch cases can be used to control the flow of logic in the program. Efficient usage of variables and data structures provides an easier solution and a lightweight, low-memory-requiring program. We have to look at the existing programming techniques, such as Divide-and-conquer, Greedy Programming, Dynamic Programming, and find out if they can be used. This problem can be solved by some basic logic or a brute-force approach. Follow the following contents to understand the approach better.

So, if the input of our problem is like n = 25, then the output will be 6

Steps

To solve this, we will follow these steps −

a := square root of n
b := cube root of n
c := cube root of a
return a + b - c

Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;
int solve(int n){
   int a = sqrt(n), b = cbrt(n), c = cbrt(a);
   return a + b - c;
}
int main(){
   int n = 25;
   cout << solve(n) << endl;
}

Input

25

Output

6