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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {
	public static void main(String[] args) {
		FastScanner sc = new FastScanner(System.in);
		int x = sc.nextInt();
		boolean c = false;
		int t = 0, i = 1;
		while(!c) {
			t += i;
			if(t >= x) {
				break;
			}
			i++;
		}
		System.out.println(i);
	}

	static class FastScanner {
		private BufferedReader reader = null;
	    private StringTokenizer tokenizer = null;
	    public FastScanner(InputStream in) {
	        reader = new BufferedReader(new InputStreamReader(in));
	        tokenizer = null;
	    }

	    public String next() {
	        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
	            try {
	                tokenizer = new StringTokenizer(reader.readLine());
	            } catch (IOException e) {
	                throw new RuntimeException(e);
	            }
	        }
	        return tokenizer.nextToken();
	    }

	    public String nextLine() {
	        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
	            try {
	                return reader.readLine();
	            } catch (IOException e) {
	                throw new RuntimeException(e);
	            }
	        }

	        return tokenizer.nextToken("\n");
	    }

	    public long nextLong() {
	        return Long.parseLong(next());
	    }

	    public int nextInt() {
	        return Integer.parseInt(next());
	    }

	    public double nextDouble() {
	         return Double.parseDouble(next());
	     }

	    public int[] nextIntArray(int n) {
	        int[] a = new int[n];
	        for (int i = 0; i < n; i++)
	            a[i] = nextInt();
	        return a;
	    }

	    public long[] nextLongArray(int n) {
	        long[] a = new long[n];
	        for (int i = 0; i < n; i++)
	            a[i] = nextLong();
	        return a;
	    }
	}
}