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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
 
#define BIG 2000000007
 
#define MOD 1000000007
typedef unsigned long long ull;
typedef   signed long long sll;
 
typedef struct {
	int a;
	int b;
} hw;
 
#define N_MAX 100000
#define M_MAX 100000
 
ull n, m, k;
ull h, w;
ull va, vb, vc, vd;
ull a[N_MAX];
// sll a[N_MAX];
ull b[N_MAX];
// ull dp[N_MAX][M_MAX + 1];
char s[N_MAX + 1];
// char t[N_MAX + 1];
// hw arr[N_MAX];
 
void swap_adj(ull *a, ull *b){
	ull tmp = *b;
	*b = *a;
	*a = tmp;
	return;
}

ull divide(ull a, ull b){
	ull x = MOD - 2;
	ull ans = 1;
	while (x) {
		if (x & 1) ans = (ans * b) % MOD;
		b = (b * b) % MOD;
		x /= 2;
	}
	return (a * ans) % MOD;
}
 
int digits(ull x){
	int i = 1;
	while (x >= 10) {
		x /= 10;
		i++;
	}
	return i;
}
 
int min(ull x, ull y){
	return (x < y) ? x : y;
}
 
ull gcd(ull x, ull y){
	if (x < y) {
		return gcd(y, x);
	} else if (y == 0) {
		return x;
	} else {
		return gcd(y, x % y);
	}
}
 
ull bitpow(ull a, ull x){
	ull result = 1;
	while (x) {
		if (x & 1) {
			result *= a;
			result %= MOD;
		}
		x /= 2;
		a = (a * a) % MOD;
	}
	return result;
}

int comp(const void *left, const void *right){
	if ((*(int*)left) < (*(int*)right)) {
		return -1;
	} else if ((*(int*)left) > (*(int*)right)) {
		return +1;
	} else {
		return 0;
	}
}

// int nextroute(int arr[]){
// 	int i = n - 1;
// 	int j, x;
// 	while (arr[i - 1] > arr[i]) i--;

// 	x = n;
// 	for (j = i; j < n; j++) {
// 		if (arr[j] < arr[i - 1]) continue;
// 		if (x == n || arr[x] > arr[j]) x = j;
// 	}
// 	arr[i - 1] ^= arr[x];
// 	arr[x] ^= arr[i - 1];
// 	arr[i - 1] ^= arr[x];

// 	qsort(&arr[i], n - i, sizeof(int), comp);
// 	return 0;
// }

int nibutan_target(ull target){
	ull maxdist = (target * (target + 1) / 2); // 時刻targetまでに到着できる距離は[-maxdist, maxdist]
	return (n <= maxdist);
}

ull solve(){
	int i, j;

	ull ok = 1000000000;
	ull ng = 0;

	// 到着できる時刻の最小値を求める
	while (ok - ng > 1) {
		ull med = (ok + ng) / 2;
		if (nibutan_target(med)) {
			ok = med;
		} else {
			ng = med;
		}
	}

	printf("%llu\n", ok);

	return 0;
}
 
int main(void){
	int i, j;
	int x;

	scanf("%llu", &n);

	solve();

	return 0;
}