問題
翻訳を表示All machines start at time 0 and work independently. A machine with time t finishes one item after t seconds, another after 2t seconds, and so on. Find the earliest whole second when their combined production reaches the target.
Implement factory_machines(times, target).
Input
times: Seconds needed for one item on each machine.
target: The number of completed items required.
Output
Return the minimum elapsed seconds.
制約
1 ≤ machines ≤ 200,000. Each time and the target are between 1 and 1,000,000,000. The answer fits signed 64-bit.
例
例 1
入力：
times = [2, 3]
target = 5
期待される出力：
expected = 6入力：
std::vector<long long> times = {2, 3};
long long target = 5;
期待される出力：
long long expected = 6;入力：
long[] times = new long[] {2, 3};
long target = 5;
期待される出力：
long expected = 6;解説
At time 6 the machines have made 3 + 2 = 5 items. At time 5 they have made only 2 + 1 = 3.
例 2
入力：
times = [4]
target = 3
期待される出力：
expected = 12入力：
std::vector<long long> times = {4};
long long target = 3;
期待される出力：
long long expected = 12;入力：
long[] times = new long[] {4};
long target = 3;
期待される出力：
long expected = 12;解説
One machine completes its third item at 3 × 4 = 12 seconds.
参考資料
この課題はCodeUtility用に調整されています。上記の説明と制約に従ってください。