Print all possible sums of consecutive numbers with sum N in C++

In this problem, we are given a positive integer N and we have to print the sequence of all possible consecutive numbers with a sum equal to N.

Let’s take an example to understand the problem,

Input: N = 15
Output: 1 2 3 4 5
7 8

A simple solution to this problem is by adding up consecutive sequence combinations till N/2. And then print the sequence that sums up to N.

Example

Live Demo

#include<iostream>
using namespace std;
void printConsequtiveSum(int N){
   int start = 1, end = (N+1)/2;
   while (start < end){
      int sum = 0;
      for (int i = start; i <= end; i++){
         sum = sum + i;
         if (sum == N){
            for (int j = start; j <= i; j++)
               cout<<j<<" ";
               cout<<endl;
               break;
         }
         if (sum > N)
            break;
      }
      sum = 0;
      start++;
   }
}
int main(){
   int N = 25;
   cout<<"Sequence of consicutive numbers that sum upto "<<N<<" are :\n";
   printConsequtiveSum(N);
   return 0;
}

Output

The sequence of consecutive numbers that sum up to 25 are −

3 4 5 6 7
12 13

This method is easy but is not so efficient.

So, we have a more complex but optimum solution that will be using a precomputed array of sums to keep track of sum. This will decrease the complexity of the sum.

Example

Live Demo

#include <iostream>
using namespace std;
void printConsequtiveSum(int N){
   int start = 1, end = 1;
   int sum = 1;
   while (start <= N/2){
      if (sum < N){
         end += 1;
         sum += end;
      }
      else if (sum > N){
         sum -= start;
         start += 1;
      }
      else if (sum == N){
         for (int i = start; i <= end; ++i)
            cout<<i<<" ";
            cout<<endl;
            sum -= start;
            start += 1;
      }
   }
}
int main(){
   int N = 25;
   cout<<"Sequence of consicutive numbers that sum upto "<<N<<" are:\n";
   printConsequtiveSum(N);
   return 0;
}