# 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;
}

## Output

Sequence of consecutive numbers that sum up to 25 are −

3 4 5 6 7
12 13