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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
#define rep(i,N) for(int i = 0; i < (N); i++)
#define erep(i,N) for(int i = N - 1; i >= 0; i--)
const ll INF = 1000000000;
const ll mod = 1000000007;
#define PI 3.1415926535
using Graph = vector<vector<int>>;
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true;} return false;}
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true;} return false;}
typedef pair<int, int> pair;
bool seen[505][505];

//dpTable
int dp[100050];

int main(){
    ll X;
    cin >> X;
    ll ans;
    rep (i,X) {
        if ((i * (i + 1) / 2) == X) {
            ans = i;
            break;
        }
        if ((i * (i + 1) / 2) < X && ((i + 1) * (i + 2) / 2) >= X) {
            ans = i + 1;
            break;
        }
    }
    cout << ans << endl;
    return 0;
}