Data type:

Unsigned Long / Double … (2^32 – 1)

Algorithm:

State of the last bulb will change if ‘I’ is a factor of N. So,

Continue (I = 1 to N) : I++

if ‘I’ is factor of N then toggle the switch

thus, we can get the last condition of the bulb.

Wow ! So simple as that. What r u waiting for ? Submit it.

Well done ! u’ll get TLE (Time Limit Exceed). Right ? I know u would be, bcoz of this silly & super bogus idea.

Lets improve it.

How many factors does N have ?

If there are Odd number of factors then output: Yes otherwise No.

A number have Odd number of factors if and only if it is a Perfect Square.

So, if (sqrt (N) = Integer) then output: Yes otherwise No.

In C/C++ we can use floor () function to judge whether a number is Integer.

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Excellent explanation