为啥新春抢红包总不是手气最佳？看完微信抢红包算法你就明白了

BigDecimal sum1 = BigDecimal.ZERO; BigDecimal redpeck ; int sum = 0; for (int i = 0; i < num.intValue(); i++) { rand[i] = random.nextInt(100); sum += rand[i]; } final BigDecimal bigDecimal = new BigDecimal(sum); BigDecimal remain = amount.subtract(min.multiply(num)); for (int i = 0; i < rand.length; i++) { if(i == num.intValue() -1){ redpeck = remain; }else{ redpeck = remain.multiply(new BigDecimal(rand[i])).divide(bigDecimal,2,RoundingMode.FLOOR); } if(remain.compareTo(redpeck)> 0){ remain = remain.subtract(redpeck); }else{ remain = BigDecimal.ZERO; } sum1= sum1.add(min.add(redpeck)).setScale(2, BigDecimal.ROUND_HALF_UP); System.out.println("第"+(i+1)+"个人抢到名片手续费为："+min.add(redpeck).setScale(2, BigDecimal.ROUND_HALF_UP)); } System.out.println("名片总额："+sum1);}// 测试标识符public static void main(String[] args) { BigDecimal amount = new BigDecimal(100).setScale(2, BigDecimal.ROUND_HALF_UP); BigDecimal min = new BigDecimal(0.01).setScale(2, BigDecimal.ROUND_HALF_UP); BigDecimal num = new BigDecimal(10).setScale(2, BigDecimal.ROUND_HALF_UP); testPocket2(amount,min,num);}激活标识符

他的渐进可谓是很高，也不是最佳选择。

演算法律条文三：割线法律条文

割线法律条文指的是把名片总手续费想象成一条更长的线段，而每个人抢到的手续费，则是这条主线段所拆分出的若干叔父线段，当所有切开点确定以后，叔父线段的间距也慢慢确定。这样每个人来抢名片的时候，只须要顺次申领与叔父线段间距等价的名片手续费均可。

private static void testPocket3(BigDecimal amount, BigDecimal min, BigDecimal num) { final Random random = new Random(); final int[] rand = new int[num.intValue()]; BigDecimal sum1 = BigDecimal.ZERO; BigDecimal redpeck; int sum = 0; for (int i = 0; i < num.intValue(); i++) { rand[i] = random.nextInt(100); sum += rand[i]; } final BigDecimal bigDecimal = new BigDecimal(sum); BigDecimal remain = amount.subtract(min.multiply(num)); for (int i = 0; i < rand.length; i++) { if (i == num.intValue() - 1) { redpeck = remain; } else { redpeck = remain.multiply(new BigDecimal(rand[i])) .divide(bigDecimal, 2, RoundingMode.FLOOR); } if (remain.compareTo(redpeck)> 0) { remain = remain.subtract(redpeck).setScale(2, BigDecimal.ROUND_HALF_UP); } else { remain = BigDecimal.ZERO; } sum1 = sum1.add(min.add(redpeck).setScale(2, BigDecimal.ROUND_HALF_UP)); System.out.println("第" + (i + 1) + "个人抢到名片手续费为：" + min.add(redpeck)); } System.out.println("名片总额：" + sum1);}// 测试标识符public static void main(String[] args) { BigDecimal amount = new BigDecimal(100).setScale(2, BigDecimal.ROUND_HALF_UP); BigDecimal min = new BigDecimal(0.01).setScale(2, BigDecimal.ROUND_HALF_UP); BigDecimal num = new BigDecimal(10).setScale(2, BigDecimal.ROUND_HALF_UP); testPocket2(amount,min,num);}激活标识符

他的渐进也相当大，但是他最致命的是效率，因为他须要进行时切开这个流程。

演算法律条文四：二倍自变量法律条文

演算法律条文四就是微信名片目前所有别于的的演算法律条文（大致简而言之，标识符模拟），二倍自变量计算表达式：2 * 只剩手续费/只剩名片数。

BigDecimal remain = amount.subtract(min.multiply(num)); final Random random = new Random(); final BigDecimal hundred = new BigDecimal("100"); final BigDecimal two = new BigDecimal("2"); BigDecimal sum = BigDecimal.ZERO; BigDecimal redpeck; for (int i = 0; i < num.intValue(); i++) { final int nextInt = random.nextInt(100); if(i == num.intValue() -1){ redpeck = remain; }else{ redpeck = new BigDecimal(nextInt).multiply(remain.multiply(two).divide(num.subtract(new BigDecimal(i)),2,RoundingMode.CEILING)).divide(hundred,2, RoundingMode.FLOOR); } if(remain.compareTo(redpeck)> 0){ remain = remain.subtract(redpeck).setScale(2, BigDecimal.ROUND_HALF_UP); }else{ remain = BigDecimal.ZERO; } sum = sum.add(min.add(redpeck)).setScale(2, BigDecimal.ROUND_HALF_UP); System.out.println("第"+(i+1)+"个人抢到名片手续费为："+min.add(redpeck)); } System.out.println("名片总额：" + sum);}激活标识符

他还是相当好的保证了每个名片手续费大致等同于，才会出现极端情况。

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