Parnic
f03184d4c4
This one's part 1 destroyed me. I had a very difficult time, trying 3 separate approaches, each one of which worked for most cases but eventually fell apart on the 5th sample or my actual puzzle input. I ended up reading a bunch of hints from the subreddit which eventually led me to a blog post describing this solution, which wasn't far off from what I had, but I was overcomplicating things. Part 2 surprised me in that I expected a simple "ore available divided by ore needed for 1 fuel" would solve it, but of course the excess chemicals produced in any given reaction meant that it wasn't that simple. So this approach uses that estimate as a lower bound, since it always underestimates, and then bisects its way to the solution (starting at the lower bound and adding 1 each time took too long). I'm sure a smarter upper bound choice could lower the runtime of this by a bit, but runtime isn't bad enough right now for me to try any additional optimizations.
50 lines
733 B
Go
50 lines
733 B
Go
package utilities
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import "math"
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func GCD[T Integer](a, b T) T {
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if b == 0 {
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return a
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}
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return GCD(b, a%b)
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}
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func LCM[T Integer](nums ...T) uint64 {
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num := len(nums)
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if num == 0 {
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return 0
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} else if num == 1 {
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return uint64(nums[0])
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}
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ret := lcm(nums[0], nums[1])
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for i := 2; i < len(nums); i++ {
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ret = lcm(uint64(nums[i]), ret)
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}
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return ret
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}
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func lcm[T Integer](a, b T) uint64 {
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return uint64(a*b) / uint64(GCD(a, b))
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}
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func Min[T Number](nums ...T) T {
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numNums := len(nums)
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if numNums == 2 {
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return T(math.Min(float64(nums[0]), float64(nums[1])))
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}
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if numNums == 0 {
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return 0
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}
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least := nums[0]
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for i := 1; i < numNums; i++ {
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if nums[i] < least {
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least = nums[i]
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}
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}
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return least
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}
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