Problem:
Score : 200 points

Problem Statement

There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0.
During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right.
That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i.
The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible.
Find the earliest possible time to reach coordinate X.

Constraints

X is an integer.

1≤X≤10^9

Input

The input is given from Standard Input in the following format:

X

Output

Print the earliest possible time for the kangaroo to reach coordinate X.

Sample Input 1

6

Sample Output 1

3

The kangaroo can reach his nest at time 3 by jumping to the right three times, which is the earliest possible time.

Sample Input 2

2

Sample Output 2

2

He can reach his nest at time 2 by staying at his position during the first second, and jumping to the right at the next second.

Sample Input 3

11

Sample Output 3

5

Solution in Kotlin (1.0.0) (judge verdict: Accepted)

fun main(arg: Array<String>) {
    val X = nextInt()
    for (t in 1 .. X) {
        if (t * (t + 1) / 2 >= X)
        return println(t)
    }
}

fun next() = readLine()!!
fun nextInt(delta: Int = 0) = Integer.parseInt(next()) + delta
fun nextLong() = next().toLong()
fun listOfString() = next().split(" ")
fun listOfInt() = listOfString().map(String::toInt)
fun listOfLong() = listOfString().map(String::toLong)
fun listOfDouble() = listOfString().map(String::toDouble)