Convert int type datetime to milliseconds

The int type datetime is expressed as follows, 13:32:53:600 => 133253600, and I wanna convert the int type datetime to the number of milliseconds that have passed since 0:00:00 of the day. For example, 133253600 => 48773600.

Consider a numeral in the form HHMMSSmmm, where HH is a decimal number of hours, MM is a decimal number of minutes SS is a decimal number of seconds, and mmm is a decimal number of milliseconds.

Interpreted as a decimal numeral of milliseconds, HH0000000 is more milliseconds than is in HH hours. There are 3,600,000 milliseconds in an hour, so 3,600,000•HH in HH hours, but HH0000000 represents 10,000,000•HH milliseconds. It has 6,400,000•HH milliseconds too many. Therefore, we can correct it by subtracting 6,400,000•HH.

Similarly, MM00000 represents 100,000•MM milliseconds, but there are only 60,000•MM milliseconds in MM minutes. It has 40,000•MM milliseconds too many, so we can correct it by subtracting 40,000•MM.

SSmmm does represent the number of milliseconds in SS seconds and mmm milliseconds, so no correction is needed for those.

We can easily find HH by dividing HHMMSSmmm by 10,000,000, using C’s truncating division. Isolating the minutes would require more work. Dividing by 100,000 would give us HHMM, and then we would need to take a remainder to separate MM. However, instead, we can rewrite the corrections to use HHMM directly.

Given some x that is HHMMSSmmm, we want x − HH•6,400,000 − MM•40,000. Adding and subtracting HH00•40,000 does not change the number, so:

• x − HH•6,400,000 − MM•40,000
• = x − HH•6,400,000 − MM•40,000 + HH00•40,000 − HH00•40,000
• = x − HH•6,400,000 + HH•100•40,000 − MM•40,000 − HH00•40,000
• = x − HH•(6,400,000 − 4,000,000) − HHMM•40,000.
• = x − HH•2,400,000 − HHMM•40,000.

As noted above, HH can be calculated in C as x/10000000, and HHMM can be calculated as x/100000.

Therefore, an expression that gives us the desired result is x - x/10000000*2400000 - x/100000*40000, which uses four multiplication/division operations and two additions/subtraction operations.

The result is calculated as follows:

r = ( ( ( H * 60 ) + M ) * 60 ) + S ) * 1000 + milli;

So if we include getting the components, that's

t = 133253600;

H = t / 10000000;
M = t / 100000 % 100;
S = t / 1000 % 100;
milli = t % 1000;

r = ( ( ( H * 60 ) + M ) * 60 ) + S ) * 1000 + milli;

But the seconds are technically already correct.

t = 133253600;

H = t / 10000000;
M = t / 100000 % 100;
milli = t % 100000;

r = ( ( H * 60 ) + M ) * 60 + milli;

That is the required 8 ops. 4 divisions, 2 multiplications and 2 additions.

You would have to divide by 10000000 (Only keeping integers) to get hours (the first 2 numbers), 100000 to get the first 4, 1000 to get the first 6, etc.
With tose numbers you then can make modulus 100 to get last 2 digits of that number and modulus 1000 to keep last 3. Last thing you have to do is transform those into miliseconds and that's it.

    int time=133253600 //example
    int hours = time/ 10000000;
    int minutes = (time/ 100000) % 100;
    int seconds = (time/ 1000) % 100;
    int milliseconds = time% 1000;

    int timeInMs = (hours * 3600000) + (minutes * 60000) + (seconds * 1000) + milliseconds;

It doesn't use 8 operations, but I can't imagine how could you do it on less. You could optimize by dividing by smaller numbers if you do one operation depending on the previous one, maybe.