还记得台球桌上的摆球三角形吗？如果需要摆满经典的5层则需要15颗球。如果拥有3颗球，则可以摆出2层结构。请按照这个规律，编写一个函数，接收一个表示球数num，返回它们可以构成的三角形层的层数。
备注：输入的球数可以大于15。如果球数不足以摆满一层的情况不算一层。
示例：
输入：1，返回：1。
输入：4，返回：2。
题目难度：简单
题目来源：CodeWars: Billiards triangle
def solution(num: int) -> int:
# pass
assert solution(1) == 1
assert solution(4) == 2
assert solution(20) == 5
assert solution(2211) == 66
assert solution(9999) == 140
def solution14(num:int):
sum=0
for i in range(0,num):
sum+=i
if num==1:
return 1
elif sum == num:
return i
elif sum > num:
return i-1
assert solution14(1) == 1
assert solution14(4) == 2
assert solution14(20) == 5
assert solution14(2211) == 66
assert solution14(9999) == 140
def solution(num: int) -> int:
n = 1
sum = 0
while True:
sum += n
if sum == num:
return n
elif sum < num:
n += 1
else:
return n-1
assert solution(1) == 1
assert solution(4) == 2
assert solution(20) == 5
assert solution(2211) == 66
assert solution(9999) == 140
def solution(num: int) -> int:
# pass
for n in range(0, num+1):
if num == ((1 + n) * n / 2):
return n
elif num < ((1 + n) * n / 2):
return n - 1
assert solution(1) == 1
assert solution(4) == 2
assert solution(20) == 5
assert solution(2211) == 66
assert solution(9999) == 140
def solution(num: int) -> int:
return 1 if num == 1 else [i - 1 for i in range(num) if i * (i + 1) > 2 * num][0]
assert solution(1) == 1
assert solution(4) == 2
assert solution(20) == 5
assert solution(2211) == 66
assert solution(9999) == 140
def sum(n):
res = 0
for i in range(1, n + 1):
res += i
return res
def solution(num: int) -> int:
layer = int((2 * num + .25) ** .5 - .5)
while layer > 0:
if sum(layer) <= num < sum(layer + 1):
return layer
else:
layer -= 1
pass
assert solution(1) == 1
assert solution(4) == 2
assert solution(20) == 5
assert solution(2211) == 66
assert solution(9999) == 140
[/quote]
def solution(num: int) -> int:
i = 1
while num-i >= 0:
num = num-i
i += 1
return i - 1
assert solution(1) == 1
assert solution(4) == 2
assert solution(20) == 5
assert solution(2211) == 66
assert solution(9999) == 140
def solution(num: int) -> int:
if num<1: return 0
level,result = 0,0
while result<num:
level+=1
result += level
return level if result == num else level-1
assert solution(1) == 1
assert solution(4) == 2
assert solution(20) == 5
assert solution(2211) == 66
assert solution(9999) == 140
assert solution(2) == 1
assert solution(0) == 0
assert solution(-1) == 0
import itertools
def solution(num: int):
for i in range(1, num+1):
if list(itertools.accumulate([j for j in range(1,i+1)]))[-1] > num:
return i-1
elif list(itertools.accumulate([j for j in range(1,i+1)]))[-1] == num:
return i
def solution(num: int) -> int:
x = 0 # 每层相加的总和
for i in range(num + 2): # 加2为了避免num为1
x = x + i
if x > num:
return i - 1
def solution(num: int) -> int:
putted_ball = 0
lines = 0
while(putted_ball<num):
putted_ball += lines + 1
if(putted_ball<=num):
lines += 1
return lines
assert solution(1) == 1
assert solution(4) == 2
assert solution(20) == 5
assert solution(2211) == 66
assert solution(9999) == 140```def solution(num: int) -> int:
num1 = 0
for i in range(num):
num1 = num1 + i + 1
if num1 - num > 0:
return i
if num1 - num == 0:
return i+1
else:
pass
assert solution(1) == 1
assert solution(4) == 2
assert solution(20) == 5
assert solution(2211) == 66
assert solution(9999) == 140
def taiqiu(num):
if num > 15:
return 5
if num < 1:
return "输入错误"
sum = 0
for i in range(1,6):
sum = sum + i
if sum > num:
return i-1
if sum == num:
return i
def solution(num: int) -> int:
import math
if num < 1:
return False
c = 2*num
# n**2 + n = 2num
# n**2 + n - 2num = 0
# 平方根为 a**2 - 4ac
value = 1 + 4*c
sq = math.sqrt(value)
x1 = (-1 + sq) / 2
return int(x1)
def solution(num: int) -> int:
n = 0
s = 0
while s < num:
n += 1
s = (n + 1) * n / 2
if s > num:
n -= 1
return n
assert solution(1) == 1
assert solution(4) == 2
assert solution(20) == 5
assert solution(2211) == 66
assert solution(9999) == 140