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>
# define rep(i, n) for(int i=0, i##_len=(n); i<i##_len; ++i)
# define reps(i, n) for(int i=1, i##_len=(n); i<=i##_len; ++i)
# define rrep(i, n) for(int i=((int)(n)-1); i>=0; --i)
# define rreps(i, n) for(int i=((int)(n)); i>0; --i)
# define ALL(x) (x).begin(), (x).end()
# define SZ(x) ((int)(x).size())
# define pb push_back
# define optimize_cin() cin.tie(0); ios::sync_with_stdio(false)
using namespace std;
using ll = long long;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
const long long INF = 1LL << 60;

ll gcd(ll a, ll b)
{
   if (a%b == 0)
   {
       return(b);
   }
   else
   {
       return(gcd(b, a%b));
   }
}

uintmax_t combination(unsigned int n, unsigned int r) {
  if ( r * 2 > n ) r = n - r;
  uintmax_t dividend = 1;
  uintmax_t divisor  = 1;
  for ( unsigned int i = 1; i <= r; ++i ) {
    dividend *= (n-i+1);
    divisor  *= i;
  }
  return dividend / divisor;
}

int modPow(long long a, long long n, long long p) {
  if (n == 1) return a % p;
  if (n % 2 == 1) return (a * modPow(a, n - 1, p)) % p;
  long long t = modPow(a, n / 2, p);
  return (t * t) % p;
}

ll lcm(ll a, ll b)
{
   return a / gcd(a, b) * b;
}

ll func(ll x){
  return (x+1)/2 * x;
}

int main()
{
  ll x;
  cin >> x;
  ll ans;

  reps(i,100000){
    if(func(i-1) < x && func(i) >= x){
      ans = i;
      break;
    }
  }
  
  cout << ans << endl;
  return 0;
}