Given a number N. The task is to print all possible sums of consecutive numbers that add up to N.

Examples:

Input : 100 Output : 9 10 11 12 13 14 15 16 18 19 20 21 22 Input :125 Output : 8 9 10 11 12 13 14 15 16 17 23 24 25 26 27 62 63

One important fact is we can not find consecutive numbers above N/2 that adds up to N, because N/2 + (N/2 + 1) would be more than N. So we start from start = 1 till end = N/2 and check for every consecutive sequence whether it adds up to N or not. If it is then we print that sequence and start looking for the next sequence by incrementing start point.

// C++ program to print consecutive sequences // that add to a given value #include<bits/stdc++.h> using namespace std; void findConsecutive(int N) { // Note that we don't ever have to sum // numbers > ceil(N/2) int start = 1, end = (N+1)/2; // Repeat the loop from bottom to half while (start < end) { // Check if there exist any sequence // from bottom to half which adds up to N int sum = 0; for (int i = start; i <= end; i++) { sum = sum + i; // If sum = N, this means consecutive // sequence exists if (sum == N) { // found consecutive numbers! print them for (int j = start; j <= i; j++) printf("%d ", j); printf("\n"); break; } // if sum increases N then it can not exist // in the consecutive sequence starting from // bottom if (sum > N) break; } sum = 0; start++; } } // Driver code int main(void) { int N = 125; findConsecutive(N); return 0; }

Output:

8 9 10 11 12 13 14 15 16 17 23 24 25 26 27 62 63

**Optimized Solution:**

In above solution, we keep recalculating sums from start to end, which results in O(N^2) worst case time complexity. This can be avoided by using a precomputed array of sums, or better yet – just keeping track of the sum you have so far and adjusting it depending on how it compares to the desired sum.

Time complexity of below code is O(N).

// Optimized C++ program to find sequences of all consecutive // numbers with sum equal to N #include <stdio.h> void printSums(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) printf("%d ", i); printf("\n"); sum -= start; start += 1; } } } // Driver Code int main() { printSums(125); return 0; }

Output:

8 9 10 11 12 13 14 15 16 17 23 24 25 26 27 62 63

**Reference : **

https://www.careercup.com/page?pid=microsoft-interview-questions&n=2

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