Home Backend Development PHP Tutorial 基于PHP代码实现中奖概率算法可用于刮刮卡、大转盘等抽奖算法_PHP

基于PHP代码实现中奖概率算法可用于刮刮卡、大转盘等抽奖算法_PHP

May 29, 2016 am 11:47 AM
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大转盘中奖概率算法在我们的日常生活中,经常遇到,那么基于php代码是如何实现中奖概率算法的,下面通过一段代码实例给大家介绍php中奖概率算法,代码简单易懂,并且附有注释,具体代码如下所示:

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<&#63;php

/*

 * 经典的概率算法,

 * $proArr是一个预先设置的数组,

 * 假设数组为:array(100,200,300,400),

 * 开始是从1,1000 这个概率范围内筛选第一个数是否在他的出现概率范围之内,

 * 如果不在,则将概率空间,也就是k的值减去刚刚的那个数字的概率空间,

 * 在本例当中就是减去100,也就是说第二个数是在1,900这个范围内筛选的。

 * 这样 筛选到最终,总会有一个数满足要求。

 * 就相当于去一个箱子里摸东西,

 * 第一个不是,第二个不是,第三个还不是,那最后一个一定是。

 * 这个算法简单,而且效率非常 高,

 * 关键是这个算法已在我们以前的项目中有应用,尤其是大数据量的项目中效率非常棒。

 */

function get_rand($proArr) {

 $result = '';

 //概率数组的总概率精度

 $proSum = array_sum($proArr);

 //概率数组循环

 foreach ($proArr as $key => $proCur) {

  $randNum = mt_rand(1, $proSum);

  if ($randNum <= $proCur) {

   $result = $key;

   break;

  } else {

   $proSum -= $proCur;

  }  

 }

 unset ($proArr);

 return $result;

}

/*

 * 奖项数组

 * 是一个二维数组,记录了所有本次抽奖的奖项信息,

 * 其中id表示中奖等级,prize表示奖品,v表示中奖概率。

 * 注意其中的v必须为整数,你可以将对应的 奖项的v设置成0,即意味着该奖项抽中的几率是0,

 * 数组中v的总和(基数),基数越大越能体现概率的准确性。

 * 本例中v的总和为100,那么平板电脑对应的 中奖概率就是1%,

 * 如果v的总和是10000,那中奖概率就是万分之一了。

 *

 */

$prize_arr = array(

 '0' => array('id'=>1,'prize'=>'平板电脑','v'=>1),

 '1' => array('id'=>2,'prize'=>'数码相机','v'=>5),

 '2' => array('id'=>3,'prize'=>'音箱设备','v'=>10),

 '3' => array('id'=>4,'prize'=>'4G优盘','v'=>12),

 '4' => array('id'=>5,'prize'=>'10Q币','v'=>22),

 '5' => array('id'=>6,'prize'=>'下次没准就能中哦','v'=>50),

);

/*

 * 每次前端页面的请求,PHP循环奖项设置数组,

 * 通过概率计算函数get_rand获取抽中的奖项id。

 * 将中奖奖品保存在数组$res['yes']中,

 * 而剩下的未中奖的信息保存在$res['no']中,

 * 最后输出json个数数据给前端页面。

 */

foreach ($prize_arr as $key => $val) {

 $arr[$val['id']] = $val['v'];

}

$rid = get_rand($arr); //根据概率获取奖项id

$res['yes'] = $prize_arr[$rid-1]['prize']; //中奖项

unset($prize_arr[$rid-1]); //将中奖项从数组中剔除,剩下未中奖项

shuffle($prize_arr); //打乱数组顺序

for($i=0;$i<count($prize_arr);$i++){

 $pr[] = $prize_arr[$i]['prize'];

}

$res['no'] = $pr;

print_r($res);

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下面再给大家分享一段实例代码基于Java实现中奖概率计算

做移动的项目,有个需求,做个摇奖的活动!其中中奖的计算比较恶心,用户要改动各个奖项的中奖概率,而且每天的奖项有个数限制。一二三四五六等奖,概率不通,怎么算一个用户参与了中没中将呢?苦思了一下,可以用Random类的 nextInt(int x)方法产生一个范围内的随机数,产生到那个区间就是几等奖了,中奖区间的产生是动态的。贴出源代码,仅供参考!

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package Mzone;

import java.util.ArrayList;

import java.util.Random;

public class Mzone {

 /**

 * CopyRright(c)2009-04:

 * Project:

 * Module ID:

 * Comments: 概率计算

 * JDK version used: <JDK1.4>

 * Author:ch

 * Create Date:2009-04-20

 * Modified By:

 * Modified Date:

 * Why & What is modified

 * Version: 1.0

 */

 static Random r = new Random();

 public static void main(String[] args) {

  //各个奖项的中奖概率的分母

  Integer _5m = new Integer(5);

  Integer _500m = new Integer(30);

  Integer _ipod = new Integer(500);

  Integer _phone = new Integer(1000);

  Integer _notebook = new Integer(1500);

  Integer _jay = new Integer(50);

  ArrayList list = new ArrayList();

  if(_5m.intValue()!=0)

   list.add(_5m);

  if(_500m.intValue()!=0)

   list.add(_500m);

  if(_ipod.intValue()!=0)

   list.add(_ipod);

  if(_phone.intValue()!=0)

   list.add(_phone);

  if(_notebook.intValue()!=0)

   list.add(_notebook);

  if(_jay.intValue()!=0)

   list.add(_jay);

  //计算最小公倍数

  int common = getN(list);

  System.out.println("最小公倍数:"+common);

  int a = 0;int b = 0;int c = 0;int d = 0;int e = 0;int f = 0;int g = 0;

  int first = 0;int second = 0;int third = 0;int four = 0;int fifth = 0;int sixth = 0;

  if(_5m.intValue()!=0){

   first = common/_5m.intValue();

  }

  if(_500m.intValue()!=0){

   second = first + (common/_500m.intValue());

  }else second = first;

  if(_ipod.intValue()!=0){

   third = second + (common/_ipod.intValue());

  }else third = second;

  if(_phone.intValue()!=0){

   four = third + (common/_phone.intValue());

  }else four = third;

  if(_notebook.intValue()!=0){

   fifth = four + (common/_notebook.intValue());

  }else fifth = four;

  if(_jay.intValue()!=0){

   sixth = fifth + (common/_jay.intValue());

  }else sixth = fifth;

  int times = 30000;//循环次数

  for(int i = 0;i < times; i++){

   int ri = getRandom(common);//产生随机数

   if(ri >= 0 && ri < first){

    a++;

   }else if(ri >= first && ri < second){

    b++;

   }else if(ri >= second && ri < third){

    c++;

   }else if(ri >= third && ri < four){

    d++;

   }else if(ri >= four && ri < fifth){

    e++;

   }else if(ri >= fifth && ri < sixth){

    f++;

   }else{

    g++;

   }

  }

  System.out.println("5m值:" + a + " 500m值:" + b + " ipodMP3:" + c + " 手机:" + d + " 笔记本电脑:" + e + " 演唱会门票:" + f + " 谢谢参与:" + g);

 }

 /**

  * 求最大公约数

 */

 public static int gcd(int m, int n){

  while (true){

   if ((m = m % n) == 0)

   return n;

   if ((n = n % m) == 0)

   return m;

  }

 }

 /**

 * 求最小公倍数

 */

 public static int gys(int z, int y){

  int t = 0;

  int c = 0;

  c = gcd(z,y);

  t = z * y / c;

  return t;

 }

 /**

  * 求几个数的最小公倍数

 */

 public static int getN(ArrayList list){

  int t = 1;

  for(int i = 0;i<list.size();i++){

   Integer temp = (Integer)list.get(i);

   t = gys(t,temp.intValue());

  }

  return t;

 }

 /**

  * 产生随机数

 */

 public static int getRandom(int y){

  int result = r.nextInt(y);

  return result;

 }

}

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