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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
using lint = long long;
bool dp[(lint(1e9 * 2))];

int main() {
  fill(begin(dp), end(dp), false);
  dp[0ll] = true;
  lint n;
  cin >> n;
  lint sum = 0ll;
  for (lint i = 0ll;; i++) {
    for (lint x = 0; x <= sum; x++) {
      if (dp[x] == true) {
        dp[x + i] = true;
        if (x > i) dp[x - i] = true;
      }
    }
    if (dp[n]) {
      cout << i << endl;
      return 0;
    }
    sum += i;
  }
}