# Count ways to express a number as sum of powers in C++

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Given two numbers num and power as input. The goal is to find the ways in which num can be represented as a sum of unique natural numbers raised to the given power. If num is 10 and power is 2 then we can represent 10 as 12+32. So total 1 way.

For Example

## Input

num=30

## Output

Count of ways to express a number as sum of powers are: 2

## Explanation

The ways in which we can express 30 as sum of powers:
12 + 22 + 52 and 12 + 22 + 32 + 42

## Input

num=35

## Output

Count of ways to express a number as sum of powers are: 1

## Explanation

The ways in which we can express ‘num’ as sum of powers: 22 + 32

Approach used in the below program is as follows

In this approach we first check if the number is itself the power of any numpower. If yes then return ways as 1, if not then recursively check for sum numpower + (num+1)power.

• Take two integers num and power as input.

• Function sum_of_powers(int num, int power, int val) takes a num and returns the count of ways to express ‘num’ as sum of unique natural numbers raised to the given power.

• Take check=(num − pow(val, power)). If check is 0 then return 1 as the number itself is valpower.

• If check is less than 0 then return 0.

• Otherwise take temp=val+1.

• Return sum of ( sum_of_powers(check, power, temp) + sum_of_powers(num, power, temp) ).

• At the end we will get ways to return value.

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
int sum_of_powers(int num, int power, int val){
int check = (num − pow(val, power));
if(check == 0){
return 1;
}
else if(check < 0){
return 0;
} else {
int temp = val + 1;
return sum_of_powers(check, power, temp) + sum_of_powers(num, power, temp);
}
}
int main(){
int num = 25, power = 2;
cout<<"Count of ways to express a number as sum of powers are: "<<sum_of_powers(num, power, 1);
return 0;
}

## Output

If we run the above code it will generate the following output −

Count of ways to express a number as sum of powers are: 2