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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

int descending_compare(const void *a, const void *b){
    if (*(int*)a > *(int*)b){
        return -1;
    }else if (*(int*)a == *(int*)b){
        return 0;
    }else{
        return 1;
    }
}

int ascending_compare(const void *a, const void *b){
    if (*(int*)a < *(int*)b){
        return -1;
    }else if (*(int*)a == *(int*)b){
        return 0;
    }else{
        return 1;
    }
}


int lower_bound(int *a, int n, int key){
    int left, mid, right;
    left = 0, right = n;
    mid = (left + right)/2;
    while ((left+1 != mid || mid+1 != right) && mid != left){
        if (key > a[mid]){
            left = mid;
        }else{
            right = mid+1;
        }
        mid = (left + right)/2;
    }
    if (a[left] >= key)return left;
    if (a[mid] >= key)return mid;
    if (a[right] >= key)return right;
    return n;
}

//greatest common divisor
unsigned long  gcd(unsigned long x, unsigned long y){
    if (y == 0){ 
        return x;
    }else if (x > y){
        return gcd(y, x % y);
    }else{
        return gcd(x, y % x);
    }
}



long long factorial(int x){
    long long rtn = 1;
    int i;
    for (i = x; i > 1; i--){
        rtn = (rtn*i);
    }
    return rtn;
}

int main(void){
    int x;
    int here = 0;
    int i;
    scanf("%d", &x);
    for (i = 1; here + i <= x; i++){
        here += i;
    }
    i += x - (here);
    printf("%d\n", i-1);

}