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Find ways an Integer can be expressed as sum of n-th power of unique natural numbers in C++
In this tutorial, we are going to write a program that find the number of ways a integer can be expressed as sum of given n-th power of unique numbers.
We have two integers number and power. And we need to find in how many ways can we represent the given number as sum of n-th power of unique natural numbers. Let's see an example.
Input − number = 50, power = 2
Output − 3
There is only possible way we can write 4 as sum of 2 powers.
We will be using recursion to solve the problem. Let's see the steps to solve the problem.
Initialize the number and power.
Write a recursive function with a suitable name. It accepts number, power and i as arguments.
If the number is less than zero or pow(i, power) is greater than number, then return 0.
If the number is zero or pow(i, power) is equal to the number, then return 1.
We have two recursive calls for the functions to compute the total number of ways
Increment the i.
In the first recursive call, check for the numbers less than given number.
In the second recursive call, check for the given number.
Example
Let's see the code.
#include <bits/stdc++.h>
using namespace std;
int findPossibleWaysCount(int number, int power, int i = 1) {
if(number < 0 || number < pow(i, power)) {
return 0;
}
if(number == 0 || number == pow(i, power)) {
return 1;
}
return findPossibleWaysCount(number - pow(i, power), power, i + 1) + findPossibleWaysCount(number, power, i + 1);
}
int main() {
// initializing the number and power
int number = 50, power = 2;
cout << findPossibleWaysCount(number, power) << endl;
return 0;
}Output
If you run the above code, then you will get the following result.
3
Conclusion
If you have any queries in the tutorial, mention them in the comment section.
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