java - 代码实现逻辑

Question

现在有100个灯泡，每个灯泡都是关着的，第一趟把所有的灯泡灯泡打开，第二趟把偶数位的灯泡制反，第三趟让第3,6,9....的灯泡制反.......第100趟让第100个灯泡制反，问经过一百趟以后有多少灯泡亮着。

代码如何实现：

Answer 1

Additional to what @hsfzxjy said

燈泡只要有一個因數就會被開關一次

• Because light switches are performed in multiples of i 趟開關的時候, 會把 i

• That is to say, the light bulb with the factor i will be switched on and off during this trip

It can be deduced from the above:

燈泡有奇數個因數最後的結果會是亮著的 (開關奇數次, 會是亮的)

It can be summarized as:

完全平方數的燈泡會亮著 (因為只有完全平方數有奇數個相異因數, 其他都會有兩兩成對的相異因數)

If you want to completely simulate this situation, here is the Python code:

lamps = [ False for i in range(100) ]

# print('starts', lamps)

for i in range(1, len(lamps)+1):
    for idx, lamp in enumerate(lamps):
        if (idx + 1) % i == 0:
            lamps[idx] = not lamp
    # print(i, lamps)

print(lamps.count(True))

But based on the above conclusion, you only need to know how many perfect square numbers there are in the number of light bulbs:

i = 1
while i**2 <= 100:
    i += 1

print(i-1)

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Questions I answered: Python-QA

Answer 2

All perfect square numbers light up