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;

int main(void) {
  int x;
  cin >> x;

  int f[1000000];
  f[0] = 0;
  f[1] = 1;
  for(int i = 2; i < 1000000; i++) {
    f[i] = f[i - 1] + i;
  }

  int tmp;
  for(int i = 0; i < 1000000; i++) {
    if(f[i] > x) {
      tmp = i - 1;
      break;
    }
    if(f[i] == x) {
      tmp = i;
      break;
    }
  }

  int diff = x - f[tmp];
  cout << tmp + diff << endl;
}