标签:locate rem money nmon 笔试题 min += 并且 分红
* 红包算法,给定一个红包总金额和分红包的人数,输出每个人随机抢到的红包数量。 * 要求: * 每个人都要抢到红包,并且金额随机 * 每个人抢到的金额数不小于1 * 每个人抢到的金额数不超过总金额的30% * 例如总金额100,人数10,输出【19 20 15 1 25 14 2 2 1 1】
//最少分得红包数 private static final double min = 1; //最多分得红包数占比 private static final double percentMax = 0.3; public void allocateMoney(double money, int peopleNum) { double minMoney = min; double maxMoney = percentMax * money; int shareMoney = 0; double sum = 0; for (int i = 0; i < peopleNum; i++) { minMoney = minMoney > money - maxMoney * (peopleNum - i - 1) ? minMoney : (money - maxMoney * (peopleNum - i - 1)); maxMoney = maxMoney < money - minMoney * (peopleNum - i - 1) ? maxMoney : (money - minMoney * (peopleNum - i - 1)); shareMoney = (int) Math.floor((maxMoney - minMoney) * Math.random() + minMoney); money = money - shareMoney; sum += shareMoney; System.out.println("第" + (i + 1) + "个人抢到:" + shareMoney + "元"); } System.out.println("要分配的红包总额为:" + sum + "元"); }
笔试题--红包算法,给定一个红包总金额和分红包的人数,输出每个人随机抢到的红包数量。
标签:locate rem money nmon 笔试题 min += 并且 分红
原文地址:https://www.cnblogs.com/cnndevelop/p/12221254.html