Good parts per run cycle
Each cycle a press makes partsPerCycle units and chips away scrapPerCycle of them. Count the good units over the whole run.
- Each cycle yields partsPerCycle minus scrapPerCycle good units.
- The good units are never negative — if scrap matches or beats output the cycle yields zero.
- The total is good units times the number of cycles.
cycleOutput(cycles: int, partsPerCycle: int, scrapPerCycle: int) → int
C++ needs a compiler and Drill does not host one yet, so this page is the reference rather than an exercise: the problem, worked examples, and the solution in full. To type it out, the same problem runs in Python.
Where you start
int cycleOutput(int cycles, int partsPerCycle, int scrapPerCycle) {
}Worked examples
| Call | Result |
|---|---|
cycleOutput(10, 5, 1) | 40 |
cycleOutput(0, 5, 1) | 0 |
cycleOutput(5, 3, 3) | 0 |
cycleOutput(5, 3, 5) | 0 |
Hint
Clamp the per-cycle good count at zero, then multiply by the cycles.
Reference solution in C++
int cycleOutput(int cycles, int partsPerCycle, int scrapPerCycle) {
int net = std::max(0, partsPerCycle - scrapPerCycle);
return cycles * net;
}