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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll INF = 1e16;
const ll mod = 1000000007;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

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;
}
int ctoi(char c) {
  if (c >= '0' && c <= '9') {
    return c - '0';
  }
  return 0;
}
ll factorial(ll n) {
  if (n == 1) {
    return 1;
  }
  ll ans = factorial(n-1);
  return ans*n;
}
ll gcd(ll a, ll b) {
  if (b == 0) return a;
  return gcd(b, a%b);
}
ll lcm(ll a, ll b) {
  return (a/gcd(a, b))*b;
}
bool is_prime(ll n) {
  for (ll i = 2; i*i <= n; i++) {
    if (n%i == 0) return false;
  }
  return n != 1;
}
ll mod_pow(ll x, ll n, ll mod) {
  if (n == 0) return 1;
  ll res = mod_pow(x*x%mod, n/2, mod);
  if (n & 1) res = res * x % mod;
  return res;
}

int main() {
  ll n;
  cin >> n;
  ll res = 0;
  ll tmp = 0;
  rep(i, n) {
    res++;
    if (n-(tmp+i+1) == 0) {
      break;
    } else if (n-(tmp+i+1) <= i+1) {
      continue;
    } else {
      tmp += i+1;
    }
  }
  cout << res << endl;
}