微信怎么抢红包(微信红包封面领取序列号)

[摘要]引言春节不可或缺的活动就是抢红包。从以前的纸质红包到现在的互联网红包(以微信红包为首)，今天我们就来分析一下抢红包的算法，其中有一些是微信红包。看完它们，你会知道最好的运气是怎么来的！

算法一:剩余量随机法不推荐算法一。算法1的全称是剩余量随机法。听名字就知道这种方法是随机分配剩余金额。我们先来看代码。

前言

春节不可或缺的活动就是抢红包。从以前的纸质红包到现在的互联网红包(以微信红包为首)，今天我们就来分析一下抢红包的算法，其中有一些是微信红包。看完它们，你会知道最好的运气是怎么来的！

算法一：剩余金额随机法

第一，算法不推荐。算法的全称是剩余量随机法。看名字就知道这种方法是随机分配剩余金额。我们先来看代码。

// 分配红包的算法private static void testPocket(BigDecimal amount, BigDecimal min, BigDecimal num) {BigDecimal remain = amount.subtract(min.multiply(num));final Random random = new Random();final BigDecimal hundred = new BigDecimal("100");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).divide(hundred, 2, RoundingMode.FLOOR); } if (remain.compareTo(redpeck) > 0) { remain = remain.subtract(redpeck); } else { remain = BigDecimal.ZERO; } sum = sum.add(min.add(redpeck)); System.out.println("第" + (i + 1) + "个人抢到红包金额为：" + min.add(redpeck).setScale(2, BigDecimal.ROUND_HALF_UP));}System.out.println("红包总额：" + sum.setScale(2, BigDecimal.ROUND_HALF_UP));}// 测试代码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);}

//分发红包私有静态void测试口袋的算法(bigdecimal amount，bigdecimal min，bigdecimal num){ bigdecimal remain = amount . subtract(min . multiple(num))；最终Random Random = new Random()；最终BigDecimal百=新BigDecimal(& # 34；100");BigDecimal sum = BigDecimal。零；BigDecimal redpeckfor(int I = 0；我& ltnum . int value()；i++){ final int nextInt = random . nextInt(100)；if(I = = num . int value()-1){ red peck = remain；} else { red peck = new BigDecimal(nextInt)。乘法(保持)。除(百，2，舍入模式。楼层)；} if(remain . compare to(red peck)& gt；0){ remain = remain . subtract(red peck)；} else { remain = BigDecimal。零；} sum = sum . add(min . add(red peck))；system . out . println(& # 34；& # 34；+(I+1)+& # 34；个人抢到的红包金额是:& # 34；+ min.add(redpeck)。setScale(2，BigDecimal。ROUND _ HALF _ UP))；} system . out . println(& # 34；红包总金额:& # 34；+ sum.setScale(2，BigDecimal。ROUND _ HALF _ UP))；}//测试代码公共静态void main(string[]args){ bigdecimal amount = newbigdecimal(100)。setscale (2，bigdecimal . round _ half _ up)；BigDecimal min =新的BigDecimal(0.01)。setScale(2，BigDecimal。ROUND _ HALF _ UP)；BigDecimal num = new BigDecimal(10)。setScale(2，BigDecimal。ROUND _ HALF _ UP)；testPocket2(amount，min，num)；}

我们可以看到，这种方法有明显的缺陷，就是一开始收到红包的人可能得到的金额最大，后来收到的金额会逐渐变小，因为他是从剩余金额中随机抽取的。显然，微信肯定不会用这种方式作为红包划分算法。不然每次有红包，都有可能马上得到最好的运气，但显然不是。

算法二：整体随机法

整额随机法的公式:红包总额*随机数/随机数之和。这个方法的核心是用一个随机数作为划分红包的标准。这个随机数是通过random类随机生成的。他的随机性比较大，看起来和我们平时抢红包的方式差不多，但是微信红包不采用这种方式，因为这种随机性太大，不公平。

private static void testPocket2(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); }else{ remain = BigDecimal.ZERO; } sum1= sum1.add(min.add(redpeck)).setScale(2, BigDecimal.ROUND_HALF_UP); System.out.println("第"+(i+1)+"个人抢到红包金额为："+min.add(redpeck).setScale(2, BigDecimal.ROUND_HALF_UP)); } System.out.println("红包总额："+sum1);}// 测试代码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);}

private static void test pocket 2(BigDecimal amount，BigDecimal min，BigDecimal num){ final Random Random = new Random()；final int[]rand = new int[num . int value()]；BigDecimal sum1 = BigDecimal。零；BigDecimal redpeckint sum = 0；for(int I = 0；我& ltnum . int value()；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；我& lt兰德.长度；i++){ if(I = = num . int value()-1){ red peck = remain；} else { red peck = remain . multiply(new BigDecimal(rand[I]))。divide(bigDecimal，2，RoundingMode。楼层)；} if(remain . compare to(red peck)& gt；0){ remain = remain . subtract(red peck)；}else{ remain = BigDecimal。零；} sum1= sum1.add(min.add(redpeck))。setScale(2，BigDecimal。ROUND _ HALF _ UP)；system . out . println(& # 34；& # 34；+(I+1)+& # 34；个人抢到的红包金额是:& # 34；+min.add(redpeck)。setScale(2，BigDecimal。ROUND _ HALF _ UP))；} system . out . println(& # 34；红包总金额:& # 34；+sum 1)；}//测试代码公共静态void main(string[]args){ bigdecimal amount = newbigdecimal(100)。setscale (2，bigdecimal . round _ half _ up)；BigDecimal min =新的BigDecimal(0.01)。setScale(2，BigDecimal。ROUND _ HALF _ UP)；BigDecimal num = new BigDecimal(10)。setScale(2，BigDecimal。ROUND _ HALF _ UP)；testPocket2(amount，min，num)；}

他的随机性很高，不是最佳选择。

算法三：割线法

割线法是指把红包的总金额想象成一条长线段，每个人抢到的金额就是这条主线段分割出来的若干子线段。当所有切割点都确定后，子线段的长度也就确定了。这样大家来抢红包的时候，只需要领取与子段长度相等的红包金额即可。

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) + "个人抢到红包金额为：" + min.add(redpeck)); } System.out.println("红包总额：" + sum1);}// 测试代码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);}

private static void test pocket 3(BigDecimal amount，BigDecimal min，BigDecimal num){ final Random Random = new Random()；final int[]rand = new int[num . int value()]；BigDecimal sum1 = BigDecimal。零；BigDecimal redpeckint sum = 0；for(int I = 0；我& ltnum . int value()；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；我& lt兰德.长度；i++){ if(I = = num . int value()-1){ red peck = remain；} else { red peck = remain . multiply(new BigDecimal(rand[I]))。divide(bigDecimal，2，RoundingMode。楼层)；} if(remain . compare to(red peck)& gt；0){ remain = remain . subtract(red peck)。setScale(2，BigDecimal。ROUND _ HALF _ UP)；} else { remain = BigDecimal。零；} sum1 = sum1.add(min.add(redpeck))。setScale(2，BigDecimal。ROUND _ HALF _ UP))；system . out . println(& # 34；& # 34；+(I+1)+& # 34；个人抢到的红包金额是:& # 34；+min . add(red peck))；} system . out . println(& # 34；红包总金额:& # 34；+sum 1)；}//测试代码公共静态void main(string[]args){ bigdecimal amount = newbigdecimal(100)。setscale (2，bigdecimal . round _ half _ up)；BigDecimal min =新的BigDecimal(0.01)。setScale(2，BigDecimal。ROUND _ HALF _ UP)；BigDecimal num = new BigDecimal(10)。setScale(2，BigDecimal。ROUND _ HALF _ UP)；testPocket2(amount，min，num)；}

他的随机性也比较大，但是他最致命的还是性能，因为他需要进行切割这一步。

算法四：二倍均值法

第四种算法是微信红包目前使用的算法(大致思路，代码模拟)，双倍平均的公式是2 *剩余金额/剩余红包数。

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)+"个人抢到红包金额为："+min.add(redpeck)); } System.out.println("红包总额：" + sum);}

BigDecimal remain = amount . subtract(min . multiply(num))；最终Random Random = new Random()；最终BigDecimal百=新BigDecimal(& # 34；100");final BigDecimal two = new BigDecimal(& # 34；2");BigDecimal sum = BigDecimal。零；BigDecimal redpeckfor(int I = 0；我& ltnum . int value()；i++){ final int nextInt = random . nextInt(100)；if(I = = num . int value()-1){ red peck = remain；} else { red peck = new BigDecimal(nextInt)。乘(保持。乘(二)。divide(num . subtract(new BigDecimal(I))，2，RoundingMode。天花板))。除(百，2，舍入模式。楼层)；} if(remain . compare to(red peck)& gt；0){ remain = remain . subtract(red peck)。setScale(2，BigDecimal。ROUND _ HALF _ UP)；}else{ remain = BigDecimal。零；} sum = sum.add(min.add(redpeck))。setScale(2，BigDecimal。ROUND _ HALF _ UP)；system . out . println(& # 34；& # 34；+(I+1)+& # 34；个人抢到的红包金额是:& # 34；+min . add(red peck))；} system . out . println(& # 34；红包总金额:& # 34；+sum)；}

他还是保证了每个红包的金额大致相等，不会出现极端的情况。