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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;
public class Main {

	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
	Scanner scan = new Scanner(System.in);
	int X = scan.nextInt();
    int sum =0;
    int d =0;
    int l =0;
    int g =0;
    for(int i=0;i<=X;i++){
    	sum += i;
		if(X-sum>i){
			d = X-sum;
		}else{
			break;
		}
    }
    for(int f =X*2/3; f>=1 ; f--){
    	sum=0;
    	for(int i=0;i<=X-f;i++){
        	sum += i;
    		if(X-f-sum>i){
    			l = X-f-sum;
    		}else{
    			break;
    		}
    }
    	if(f>l){
    		g =f;
    	}else{
    		break;
    	}
 }
    if(g<d){
 	   d=g;
    }
    System.out.println(d);
}
}