高效优化的算法
高效优化的算法
提问于 2019-07-12 02:28:28
我正在尝试制定优化每日任务计划的算法。例如：
1. There are 16 tasks to finish in one day, expirience worker can finish 8 tasks and newer worker could finish 4 tasks. As output we should get 2 expirienced worker and none new worker.现在我有这段代码，但它不能正常工作：
private void calculate(int numberOfTasks, int oldWorkerValue, int youngWorkerValue) {
int numberOfOldWorkers = numberOfTasks / oldWorkerValue;
int numberOfYoungWorkers = 0;
if(numberOfOldWorkers == 0) {
numberOfOldWorkers = 1;
} else {
numberOfYoungWorkers = (numberOfTasks % numberOfOldWorkers) / youngWorkerValue;
if(numberOfTasks % oldWorkerValue != youngWorkerValue*numberOfYoungWorkers) {
numberOfYoungWorkers = numberOfYoungWorkers + 1;
}
}
System.out.println("Expirienced worker: " + numberOfOldWorkers + " and new workers: " + numberOfYoungWorkers);
}它返回了工人的数量，但它并不像它应该的那样工作。我没有像我在这几个例子中写的那样得到结果。你能给我一些建议吗？
回答 1
Stack Overflow用户
回答已采纳
发布于 2019-07-12 02:59:54
我不确定你是否能找到确切的公式，所以我会做简单的蛮力：
public static Pair<Integer, Integer> calculate(int numberOfTasks, int oldWorkerValue, int youngWorkerValue) {
int m = Integer.MAX_VALUE; // minimal value
int o = 0; // old workers
int n = 0; // new workers
for (int i = 1; i <= (int)Math.ceil((double) numberOfTasks / oldWorkerValue); i++) {
int nw = (int) Math.ceil((double)(numberOfTasks - i * oldWorkerValue) / youngWorkerValue);
if (nw < 0) nw = 0;
int tm = i * oldWorkerValue + nw * youngWorkerValue;
if (tm < m) {
m = tm; o = i; n = nw;
}
}
return new Pair<>(o, n); // pair <old workers, new workers>
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/56995322
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