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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <limits.h>

int min = INT_MAX;
int x;

void Solve(int cx,int t);
int main(int argc, char const *argv[])
{

	int cx = 0;
	scanf("%d",&x);
	min = x;
	//Solve(0,0);
	//printf("%d\n",min );
	int i;
	for(i=0;i<=x;i++){
		if((i*(i+1))/2 >= x){
			printf("%d\n",i );
		}
	}
	return 0;
}
/*void Solve(int cx,int t){
	if(t > min){
		return;
	}
	else if(cx == x){
		if(t < min){
			min = t;
		}
		return;
	}
	else{
		Solve(cx,t+1);
		Solve(cx+(t+1),t+1);
		// Solve(cx-(t+1),t+1);
	}
}*/