Print all integers that are sum of powers of two given numbers in C++

In this problem, we are given two numbers a and b and an integer bound and we have to print all values less than binding which is the sum of squares of a and b.

Bound >= ai + bj

Let’s take an example to understand the problem −

Input: a=2, b=3, bound=8
Output: 2 3 4 5 7

To solve this problem, we will use nested loops using two variables i and j from 0. The outer loop will have ending condition xi = bound and the inner loop will have ending condition xi + yj > bound. For each iteration of the inner loop, we will store the value of xi + yi in a sorted list contains all such values. And then at the end print all values of the list.

Example

Program to show the implementation of our solution −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
void powerSum(int x, int y, int bound) {
   set<int> sumOfPowers;
   vector<int> powY;
   int i;
   powY.push_back(1);
   for (i = y; i < bound; i = i * y)
      powY.push_back(i);
   i = 0;
   while (true) {
      int powX = pow(x, i);
      if (powX >= bound)
         break;
      for (auto j = powY.begin(); j != powY.end(); ++j) {
         int num = powX + *j;
         if (num <= bound)
            sumOfPowers.insert(num);
         else
            break;
      }
      i++;
   }
   set<int>::iterator itr;
   for (itr = sumOfPowers.begin(); itr != sumOfPowers.end(); itr++) {
      cout<<*itr <<" ";
   }
}
int main() {
   int x = 2, y = 3, bound = 25;
   cout<<"Sum of powers of "<<x<<" and "<<y<<" less than "<<bound<<" are :\n";
   powerSum(x, y, bound);
   return 0;
}

Output

Sum of powers of 2 and 3 less than 25 are −
2 3 4 5 7 9 10 11 13 17 19 25