Beginner · loops + accumulator

Sum from 1 to N

Read n and print 1 + 2 + … + n. The accumulator pattern in its purest form — and a chance to meet Gauss's shortcut.

Problem

Input: one integer n (1 ≤ n ≤ 10,000).

Output: a single line with the sum of all integers from 1 to n.

Sample:

Input	Output
`5`	`15`

Your solution

Program.cs — your solution

Custom input for "Run my input"

Judging runs entirely in your browser. Your code is saved locally as you type.

Stuck? Hints

Hint 1 — the accumulator

Declare int sum = 0; before the loop. Each pass does sum += i; — that's the pattern you'll reuse in every aggregation you ever write.

Hint 2 — or skip the loop entirely

Gauss: the sum of 1..n is n * (n + 1) / 2. Try proving to yourself why n * (n + 1) is always even.

Solution (try first!)

int n = int.Parse(Console.ReadLine());
int sum = 0;
for (int i = 1; i <= n; i++)
{
    sum += i;
}
Console.WriteLine(sum);

// Gauss's formula — O(1), no loop
int n = int.Parse(Console.ReadLine());
Console.WriteLine(n * (n + 1) / 2);

Watch the integer division in the formula version: n * (n + 1) / 2 divides the product, which is always even — so no fraction is ever lost. Next up: Maximum of Three.