【每日一题0104】台球金字塔

:woman_mage: 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。

image

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
2 Likes
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
1 Like
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
1 Like
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

image

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