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)

/** "Ever tried.
         Ever failed.
               No matter.
                   Try Again.
                        Fail again.
                             Fail better.**/
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define ll long long
#define bp  __builtin_popcount
#define pb push_back
#define loop(i,n) for(int i = 0;i < int(n);i++)
#define sz(s) (int)(s.size())
#define all(x) x.begin(), x.end()
#define allr(x) x.rbegin(), x.rend()
#define setdp memset(dp,-1,sizeof(dp))
#define fr first
#define sc second
#define NumofDigits(n)   ((long long)log10(n)+1)
using namespace std;
using namespace __gnu_pbds;
typedef tree<int,null_type,less<int>,rb_tree_tag, tree_order_statistics_node_update> indexed_set;
const int maxn=1e5+5;
const  double EPS = 1e-9;
const int MOD = 1e9+7;
/*
ll gcd(ll a, ll b) { return !b ? a : gcd(b, a % b); }
ll lcm(ll a, ll b) { return (a / gcd(a, b)) * b; }
ll fastpow(ll b, ll p)
{if(!p)return 1;ll ret = fastpow(b, p >> 1);ret *= ret;if(p&1)ret*= b;return ret;}
int dx[] = {- 1, 0, 0, 1, 0 };
int dy[] = { 0,- 1, 1, 0, 0 };*/
void File()
{
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
}

int main()
{
    ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    ll n;
    cin>>n;
    ll ans=0;
   for(ll i=1;i<=n;i++)
   {
       ans+=i;
       if(ans>=n)
         return cout<<i,0;
   }
}