The [Floyd's triangle](https://en.wikipedia.org/wiki/Floyd%27s_triangle) is a right-angled triangular array of natural numbers listing them in order, in lines of increasing length, so a Floyds triangle of size 6 looks like:

```
1
2  3
4  5  6
7  8  9  10
11 12 13 14 15
16 17 18 19 20 21
    ...
```

In this kata you're given a number, and expected to return the line number it falls in, in the Floyd's triangle

```python
nth_floyd(3) # => 2 (i.e the number `3` falls in line 2 of the triangle)
nth_floyd(17) # => 6

nth_floyd(22) # => 7
nth_floyd(499502) # => 1000
```

```python
nth_floyd(n)
```

## Constraints

### 1 <= n <= 10^(9)

Difficulty: introductory

Starter code:

def nth_floyd(n):
	

Solutions:
def nth_floyd(n):
  return ((1+8*(n-1))**0.5+1)//2

from math import ceil

def nth_floyd(n):
    return ceil(((8*n + 1) ** 0.5 - 1) / 2)

def nth_floyd(n):
  s=i=0
  while s<n:
      i+=1
      s+=i
  return i

def nth_floyd(n):
  return round((n * 2)**(.5))

from math import sqrt

def nth_floyd(n):
    return round(sqrt(n * 2))

def nth_floyd(n):
  sum = 0
  for x in range(1,n):
    if sum + x >= n:
        return x
    sum += x

def nth_floyd(n):
    return int((2*n)**.5+.5)

from math import ceil
def nth_floyd(t):
    return ceil((-1+(1+8*t)**0.5)/2)

def nth_floyd(n):
    i = 0
    j = 0
    while n > j:
        i += 1
        j += i
    return i

def nth_floyd(n):
    a = (n * 8 + 1) ** 0.5
    return a // 2 if a % 1 <= 1e-10 else round(a / 2)