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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#define _USE_MATH_DEFINES
#include <math.h>
#include <iomanip>
#include <stdio.h>
#include <numeric>
#include <cmath>
#include <limits>
#include <bitset> 
#include <iterator> 
#include <map>
#include <queue>
#include <set>
#define rep(i,n) for(ll i=0;i<(n); ++i)
#define per(i,n) for(ll i=n-1;i>=0; --i)
#define all(x) (x).begin(),(x).end()
using namespace std;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const long long INF = 1LL << 60;
typedef long long ll;
ll MOD = 1000000007;
ll gcd(ll a, ll b) {
	if (b == 0) return a;
	return gcd(b, a % b);
}

int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	ll x;
	cin >> x;
	cout << ceil((-1 + sqrt(1 + 8 * x)) / 2.0) << endl;
}